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data/3.算法高阶/1.leetcode-树/10.103-二叉树的锯齿形层序遍历/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-1457c70ac29a46a795c3a8aa8de0d34f",
"keywords": [
"leetcode",
"二叉树的锯齿形层序遍历"
],
"children": [],
"export": [
"solution.json"
],
"title": "二叉树的锯齿形层序遍历"
}
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/10.103-二叉树的锯齿形层序遍历/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p>给定一个二叉树，返回其节点值的锯齿形层序遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。</p>
<p>例如：<br />
给定二叉树 <code>[3,9,20,null,null,15,7]</code>,</p>
<pre>
3
/ \
9 20
/ \
15 7
</pre>
<p>返回锯齿形层序遍历如下：</p>
<pre>
[
[3],
[20,9],
[15,7]
]
</pre>
data/3.算法高阶/1.leetcode-树/10.103-二叉树的锯齿形层序遍历/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
#include <vector>
#include <queue>
#include <cstddef>
using std::vector;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
vector<vector<int>> retv;
if (!root)
return retv;
std::queue<TreeNode *> q;
q.push(root);
bool reverse = false;
vector<int> v;
for (int cur_level_sz = 1, next_level_sz = 0; !q.empty();)
{
TreeNode *node = q.front();
q.pop();
v.push_back(node->val);
--cur_level_sz;
if (node->left)
{
q.push(node->left);
++next_level_sz;
}
if (node->right)
{
q.push(node->right);
++next_level_sz;
}
if (!cur_level_sz)
{
retv.push_back(reverse ? vector<int>(v.crbegin(), v.crend()) : v);
v.clear();
cur_level_sz = next_level_sz;
next_level_sz = 0;
reverse = !reverse;
}
}
return retv;
}
};
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/10.103-二叉树的锯齿形层序遍历/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d55fa143ea4b4a9b9cfc24a1e68e0e64"
}
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/10.103-二叉树的锯齿形层序遍历/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 二叉树的锯齿形层序遍历
<p>给定一个二叉树，返回其节点值的锯齿形层序遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。</p>
<p>例如：<br />
给定二叉树 <code>[3,9,20,null,null,15,7]</code>,</p>
<pre>
3
/ \
9 20
/ \
15 7
</pre>
<p>返回锯齿形层序遍历如下：</p>
<pre>
[
[3],
[20,9],
[15,7]
]
</pre>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
```
## 答案
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
vector<vector<int>> result;
if (root == NULL)
{
return result;
}
deque<TreeNode *> node_queue;
node_queue.push_back(root);
node_queue.push_back(NULL);
deque<int> level_list;
bool is_order_left = true;
while (node_queue.size() > 0)
{
TreeNode *curr_node = node_queue.front();
node_queue.pop_front();
if (curr_node != NULL)
{
if (curr_node->left != NULL)
{
node_queue.push_back(curr_node->left);
}
if (curr_node->right != NULL)
{
node_queue.push_back(curr_node->right);
}
}
else
{
vector<int> tmp;
for (int a : level_list)
{
tmp.push_back(a);
}
result.push_back(tmp);
level_list.clear();
if (node_queue.size() > 0)
{
node_queue.push_back(NULL);
}
is_order_left = !is_order_left;
}
}
return result;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
if (root == NULL)
return {};
vector<vector<int>> res;
int flag = 1;
stack<TreeNode *> a, b;
a.push(root);
while (!a.empty() || !b.empty())
{
vector<int> cur;
while (flag == 1 && !a.empty())
{
root = a.top();
cur.push_back(root->val);
a.pop();
if (root->left != NULL)
b.push(root->left);
if (root->right != NULL)
b.push(root->right);
}
while (flag == -1 && !b.empty())
{
root = b.top();
cur.push_back(root->val);
b.pop();
if (root->right != NULL)
a.push(root->right);
if (root->left != NULL)
a.push(root->left);
}
flag = -1 * flag;
res.push_back(cur);
}
return res;
}
};
```
### B
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
vector<vector<int>> ans;
if (!root)
return ans;
queue<TreeNode *> qnode;
bool orderByLeft = true;
qnode.push(root);
while (!qnode.empty())
{
int levelSize = qnode.size();
deque<int> level;
for (int i = 0; i < levelSize; i++)
{
auto curNode = qnode.front();
qnode.pop();
if (orderByLeft)
{
level.push_back(curNode->val);
}
else
{
level.push_front(curNode->val);
}
if (curNode->left != NULL)
qnode.push(curNode->left);
if (curNode->right != NULL)
qnode.push(curNode->right);
}
orderByLeft = !orderByLeft;
vector<int> curlevel{level.begin(), level.end()};
ans.push_back(curlevel);
}
return ans;
}
};
```
### C
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
if (!root)
return {};
vector<vector<int>> res;
queue<TreeNode *> q{{root}};
int cnt = 0;
while (!q.empty())
{
vector<int> oneLevel;
for (int i = q.size(); i > 0; i--)
{
TreeNode *t = q.front();
q.pop();
oneLevel.push_back(t->val);
if (t->left)
q.push(t->left);
if (t->right)
q.push(t->right);
}
if (cnt % 2 == 1)
reverse(oneLevel.begin(), oneLevel.end());
res.push_back(oneLevel);
cnt++;
}
return res;
}
};
```
### D
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
vector<vector<int>> results;
if (root == NULL)
{
return results;
}
zigzagLevelOrderDFS_helper(root, 0, results);
return results;
}
void zigzagLevelOrderDFS_helper(TreeNode *node, int level, vector<vector<int>> &results)
{
if (level >= results.size())
{
vector<int> new_level;
new_level.push_back(node->val);
results.push_back(new_level);
}
else
{
if (level % 2 == 0)
{
results[level].push_back(node->val);
}
else
{
results[level].insert(results[level].begin(), node->val);
}
}
if (node->left != NULL)
{
zigzagLevelOrderDFS_helper(node->left, level + 1, results);
}
if (node->right != NULL)
{
zigzagLevelOrderDFS_helper(node->right, level + 1, results);
}
}
};
```
data/3.算法高阶/1.leetcode-树/8.101-对称二叉树/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-587f3b68dc7047d5b2445cb0c5fc50c3",
"keywords": [
"leetcode",
"对称二叉树"
],
"children": [],
"export": [
"solution.json"
],
"title": "对称二叉树"
}
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/8.101-对称二叉树/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p>给定一个二叉树，检查它是否是镜像对称的。</p>
<p>&nbsp;</p>
<p>例如，二叉树&nbsp;<code>[1,2,2,3,4,4,3]</code> 是对称的。</p>
<pre> 1
/ \
2 2
/ \ / \
3 4 4 3
</pre>
<p>&nbsp;</p>
<p>但是下面这个&nbsp;<code>[1,2,2,null,3,null,3]</code> 则不是镜像对称的:</p>
<pre> 1
/ \
2 2
\ \
3 3
</pre>
<p>&nbsp;</p>
<p><strong>进阶：</strong></p>
<p>你可以运用递归和迭代两种方法解决这个问题吗？</p>
data/3.算法高阶/1.leetcode-树/8.101-对称二叉树/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
#include <cstddef>
#include <stack>
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
if (root == NULL)
return true;
TreeNode *lt = root->left, *rt = root->right;
for (std::stack<TreeNode *> stack; !stack.empty() || (lt && rt);)
{
if (lt && rt)
{
if (lt->val != rt->val)
return false;
stack.push(lt->right);
stack.push(rt->left);
lt = lt->left;
rt = rt->right;
}
else if (lt || rt)
return false;
else
{
lt = stack.top();
stack.pop();
rt = stack.top();
stack.pop();
}
}
if (lt || rt)
return false;
else
return true;
}
};
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/8.101-对称二叉树/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f1ce32e09e1e4a6baa582a6432191e67"
}
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/8.101-对称二叉树/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 对称二叉树
<p>给定一个二叉树，检查它是否是镜像对称的。</p>
<p>&nbsp;</p>
<p>例如，二叉树&nbsp;<code>[1,2,2,3,4,4,3]</code> 是对称的。</p>
<pre> 1
/ \
2 2
/ \ / \
3 4 4 3
</pre>
<p>&nbsp;</p>
<p>但是下面这个&nbsp;<code>[1,2,2,null,3,null,3]</code> 则不是镜像对称的:</p>
<pre> 1
/ \
2 2
\ \
3 3
</pre>
<p>&nbsp;</p>
<p><strong>进阶：</strong></p>
<p>你可以运用递归和迭代两种方法解决这个问题吗？</p>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
```
## 答案
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
if (!root)
return true;
queue<TreeNode *> q;
q.push(root->left);
q.push(root->right);
while (!q.empty())
{
auto p1 = q.front();
q.pop();
auto p2 = q.front();
q.pop();
if (!p1 && !p2)
continue;
if (!p1 || !p2)
return false;
if (p1->val != p2->val)
return false;
q.push(p1->right);
q.push(p1->left);
q.push(p2->right);
q.push(p2->left);
}
return true;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
queue<TreeNode *> q1, q2;
if (!root)
return true;
q1.push(root->left);
q2.push(root->right);
while (!q1.empty() && !q2.empty())
{
TreeNode *node1 = q1.front();
q1.pop();
TreeNode *node2 = q2.front();
q2.pop();
if (!node1 && !node2)
continue;
if ((!node1 && node2) || (node1 && !node2) || (node1->val != node2->val))
return false;
q1.push(node1->left);
q1.push(node1->right);
q2.push(node2->right);
q2.push(node2->left);
}
return true;
}
};
```
### B
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
if (!root)
return true;
return helper(root->left, root->right);
}
bool helper(TreeNode *left, TreeNode *right)
{
if (!left && !right)
return true;
if ((!left && right) || (left && !right) || (left->val != right->val))
return false;
return helper(left->left, right->right) && helper(left->right, right->left);
}
};
```
### C
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
return ismirror(root, root);
}
bool ismirror(TreeNode *t1, TreeNode *t2)
{
if (t1 == NULL && t2 == NULL)
return true;
if (t1 == NULL || t2 == NULL)
return false;
return (t1->val == t2->val) && ismirror(t1->left, t2->right) && ismirror(t1->right, t2->left);
}
};
```
data/3.算法高阶/1.leetcode-树/9.102-二叉树的层序遍历/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-894e8b4e12a64b228b73d75b3fe22817",
"keywords": [
"leetcode",
"二叉树的层序遍历"
],
"children": [],
"export": [
"solution.json"
],
"title": "二叉树的层序遍历"
}
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/9.102-二叉树的层序遍历/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p>给你一个二叉树，请你返回其按 <strong>层序遍历</strong> 得到的节点值。 （即逐层地，从左到右访问所有节点）。</p>
<p> </p>
<p><strong>示例：</strong><br />
二叉树：<code>[3,9,20,null,null,15,7]</code>,</p>
<pre>
3
/ \
9 20
/ \
15 7
</pre>
<p>返回其层序遍历结果：</p>
<pre>
[
[3],
[9,20],
[15,7]
]
</pre>
data/3.算法高阶/1.leetcode-树/9.102-二叉树的层序遍历/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
#include <vector>
#include <deque>
#include <cstddef>
using std::deque;
using std::vector;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
vector<vector<int>> levelOrder(TreeNode *root)
{
vector<vector<int>> retv;
if (root)
{
deque<TreeNode *> q{root};
while (!q.empty())
{
vector<int> v;
for (decltype(q.size()) i = 0, n = q.size(); i != n; ++i)
{
TreeNode *node = q.front();
q.pop_front();
v.push_back(node->val);
if (node->left)
q.push_back(node->left);
if (node->right)
q.push_back(node->right);
}
retv.push_back(v);
}
}
return retv;
}
};
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/9.102-二叉树的层序遍历/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "9290e1345143424f8a116ab56f88f50e"
}
\ No newline at end of file
data/3.算法高阶/1.leetcode-树/9.102-二叉树的层序遍历/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 二叉树的层序遍历
<p>给你一个二叉树，请你返回其按 <strong>层序遍历</strong> 得到的节点值。 （即逐层地，从左到右访问所有节点）。</p>
<p> </p>
<p><strong>示例：</strong><br />
二叉树：<code>[3,9,20,null,null,15,7]</code>,</p>
<pre>
3
/ \
9 20
/ \
15 7
</pre>
<p>返回其层序遍历结果：</p>
<pre>
[
[3],
[9,20],
[15,7]
]
</pre>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
```
## 答案
```cpp
class Solution
{
public:
vector<vector<int>> levelOrder(TreeNode *root)
{
vector<vector<int>> res;
queue<TreeNode *> q;
if (root != NULL)
{
while (!q.empty())
{
int size = q.size();
vector<int> temp;
for (int i = 0; i < size; i++)
{
TreeNode *t = q.front();
q.pop();
temp.push_back(t->val);
if (t->left != NULL)
q.push(t->left);
if (t->right != NULL)
q.push(t->right);
}
res.push_back(temp);
temp.clear();
}
}
return res;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
vector<vector<int>> levelOrder(TreeNode *root)
{
vector<vector<int>> ret;
if (!root)
return ret;
queue<TreeNode *> q;
q.push(root);
while (!q.empty())
{
int currNodeSize = q.size();
ret.push_back(vector<int>());
for (int i = 1; i <= currNodeSize; i++)
{
TreeNode *node = q.front();
q.pop();
ret.back().push_back(node->val);
if (node->left)
q.push(node->left);
if (node->right)
q.push(node->right);
}
}
return ret;
}
};
```
### B
```cpp
class Solution
{
public:
int depth(TreeNode *root)
{
if (root == NULL)
return 0;
return max(depth(root->left), depth(root->right)) + 1;
}
void levelOrder(vector<vector<int>> &ans, TreeNode *node, int level)
{
if (!node)
return;
ans[level].push_back(node->val);
levelOrder(ans, node->left, level - 1);
levelOrder(ans, node->right, level - 1);
}
vector<vector<int>> levelOrderBottom(TreeNode *root)
{
int d = depth(root);
vector<vector<int>> ans(d, vector<int>{});
levelOrder(ans, root, d - 1);
return ans;
}
};
```
### C
```cpp
class Solution
{
public:
void dfs(TreeNode *root, vector<vector<int>> &res)
{
queue<TreeNode *> q;
q.push(root);
while (!q.empty())
{
int sz = q.size();
vector<int> tp;
while (sz > 0)
{
TreeNode *p = q.front();
q.pop();
if (p->left != NULL)
{
q.push(p->left);
}
if (p->right != NULL)
{
q.push(p->right);
}
tp.push_back(p->val);
sz--;
}
res.push_back(tp);
}
}
vector<vector<int>> levelOrder(TreeNode *root)
{
vector<vector<int>> res;
if (root == NULL)
{
return res;
}
dfs(root, res);
return res;
}
};
```
data/3.算法高阶/2.leetcode-哈希表/10.166-分数到小数/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-602c74771c664b6798149723fa790697",
"keywords": [
"leetcode",
"分数到小数"
],
"children": [],
"export": [
"solution.json"
],
"title": "分数到小数"
}
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/10.166-分数到小数/desc.html 0 → 100644
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<p>给定两个整数，分别表示分数的分子 <code>numerator</code> 和分母 <code>denominator</code>，以 <strong>字符串形式返回小数</strong> 。</p>
<p>如果小数部分为循环小数，则将循环的部分括在括号内。</p>
<p class="MachineTrans-lang-zh-CN">如果存在多个答案，只需返回 <strong>任意一个</strong> 。</p>
<p class="MachineTrans-lang-zh-CN">对于所有给定的输入，<strong>保证</strong> 答案字符串的长度小于 <code>10<sup>4</sup></code> 。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>numerator = 1, denominator = 2
<strong>输出：</strong>"0.5"
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>numerator = 2, denominator = 1
<strong>输出：</strong>"2"
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>numerator = 2, denominator = 3
<strong>输出：</strong>"0.(6)"
</pre>
<p><strong>示例 4：</strong></p>
<pre>
<strong>输入：</strong>numerator = 4, denominator = 333
<strong>输出：</strong>"0.(012)"
</pre>
<p><strong>示例 5：</strong></p>
<pre>
<strong>输入：</strong>numerator = 1, denominator = 5
<strong>输出：</strong>"0.2"
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>-2<sup>31</sup> <= numerator, denominator <= 2<sup>31</sup> - 1</code></li>
<li><code>denominator != 0</code></li>
</ul>
data/3.算法高阶/2.leetcode-哈希表/10.166-分数到小数/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
int m1 = numerator > 0 ? 1 : -1;
int m2 = denominator > 0 ? 1 : -1;
long long num = abs((long long)numerator);
long long den = abs((long long)denominator);
long long t = num / den;
long long res = num % den;
string ans = to_string(t);
if (m1 * m2 == -1 && (t > 0 || res > 0))
ans = '-' + ans;
if (res == 0)
return ans;
ans += '.';
unordered_map<long long, int> m;
string s = "";
int pos = 0;
while (res != 0)
{
if (m.find(res) != m.end())
{
s.insert(m[res], "(");
s += ')';
return ans + s;
}
m[res] = pos;
s += to_string((res * 10) / den);
res = (res * 10) % den;
pos++;
}
return ans + s;
}
};
data/3.算法高阶/2.leetcode-哈希表/10.166-分数到小数/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "1a7236d672a24076bcc0f1eff7050847"
}
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data/3.算法高阶/2.leetcode-哈希表/10.166-分数到小数/solution.md 0 → 100644
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# 分数到小数
<p>给定两个整数，分别表示分数的分子 <code>numerator</code> 和分母 <code>denominator</code>，以 <strong>字符串形式返回小数</strong> 。</p>
<p>如果小数部分为循环小数，则将循环的部分括在括号内。</p>
<p class="MachineTrans-lang-zh-CN">如果存在多个答案，只需返回 <strong>任意一个</strong> 。</p>
<p class="MachineTrans-lang-zh-CN">对于所有给定的输入，<strong>保证</strong> 答案字符串的长度小于 <code>10<sup>4</sup></code> 。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>numerator = 1, denominator = 2
<strong>输出：</strong>"0.5"
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>numerator = 2, denominator = 1
<strong>输出：</strong>"2"
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>numerator = 2, denominator = 3
<strong>输出：</strong>"0.(6)"
</pre>
<p><strong>示例 4：</strong></p>
<pre>
<strong>输入：</strong>numerator = 4, denominator = 333
<strong>输出：</strong>"0.(012)"
</pre>
<p><strong>示例 5：</strong></p>
<pre>
<strong>输入：</strong>numerator = 1, denominator = 5
<strong>输出：</strong>"0.2"
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>-2<sup>31</sup> <= numerator, denominator <= 2<sup>31</sup> - 1</code></li>
<li><code>denominator != 0</code></li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
string res;
int numerator = 1, denominator = 2;
res = sol.fractionToDecimal(numerator, denominator);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
const static int CACHE_SIZE = 1000;
string fractionToDecimal(int numerator, int denominator)
{
if (numerator == 0)
return "0";
bool isNegative = numerator < 0 ^ denominator < 0;
double doubleNumerator = fabs(numerator);
double doubleDenominator = fabs(denominator);
double integral = floor(doubleNumerator / doubleDenominator);
doubleNumerator -= doubleDenominator * integral;
int len = 0;
string cache;
vector<double> dividendCache;
bool isRecurring = false;
int recurringStart = 0;
int i = 0;
while (doubleNumerator)
{
doubleNumerator *= 10;
len = dividendCache.size();
for (i = 0; i < len; ++i)
if (dividendCache[i] == doubleNumerator)
{
isRecurring = true;
recurringStart = i;
break;
}
if (isRecurring)
break;
i = (int)(floor(doubleNumerator / doubleDenominator));
cache += i;
dividendCache.push_back(doubleNumerator);
doubleNumerator -= doubleDenominator * i;
}
string ret = isNegative ? "-" : "";
string tmp;
char c;
if (integral == 0)
ret += "0";
else
{
while (integral)
{
c = (int)(integral - 10 * floor(integral / 10)) + '0';
tmp = c + tmp;
integral = floor(integral / 10);
}
ret += tmp;
}
if (dividendCache.size() > 0)
ret += '.';
i = 0;
if (isRecurring)
{
ret += cache.substr(0, recurringStart);
ret += '(';
ret += cache.substr(recurringStart);
ret += ')';
}
else
ret += cache;
return ret;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
typedef long long LL;
LL x = numerator, y = denominator;
if (x % y == 0)
return to_string(x / y);
string res;
if ((x < 0) ^ (y < 0))
res += '-';
x = abs(x), y = abs(y);
res += to_string(x / y) + '.';
x %= y;
unordered_map<LL, int> hash;
while (x)
{
hash[x] = res.size();
x *= 10;
res += to_string(x / y);
x %= y;
if (hash.count(x))
{
res = res.substr(0, hash[x]) + '(' + res.substr(hash[x]) + ')';
break;
}
}
return res;
}
};
```
### B
```cpp
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
string ans;
unordered_map<long, int> use;
if (numerator > 0 && denominator < 0 || numerator < 0 && denominator > 0)
ans = '-';
long numerat = fabs(numerator);
long denominat = fabs(denominator);
ans += to_string(numerat / denominat);
numerat = numerat % denominat;
int point, x;
if (numerat)
{
ans += ".";
point = ans.length() - 1;
}
else
return ans;
while (numerat && use.count(numerat) == 0)
{
use[numerat] = ++point;
numerat *= 10;
x = numerat / denominat;
numerat = numerat % denominat;
ans.push_back(x + '0');
}
if (numerat)
{
ans.insert(ans.begin() + use[numerat], '(');
ans.push_back(')');
}
return ans;
}
};
```
### C
```cpp
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
int m1 = numerator > 0 ? 1 : -1;
int m2 = denominator > 0 ? 1 : -1;
long long num = abs((long long)numerator);
long long den = abs((long long)denominator);
long long t = num / den;
long long res = num % den;
string ans = to_string(t);
if (m1 * m2 == -1 && (t > 0 || res > 0))
ans = '-' + ans;
if (res == 0)
return ans;
ans += '.';
unordered_map<long long, int> m;
string s = "";
int pos = 0;
while (res != 0)
{
if (m.find(res) != m.end())
{
s.insert(m[res], "(");
s += ')';
return ans + s;
}
m[res] = pos;
s += to_string((res * 10) / den);
res = (res * 10) % den;
pos++;
}
return ans + s;
}
};
```
data/3.算法高阶/2.leetcode-哈希表/7.126-单词接龙 II/config.json 0 → 100644
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{
"node_id": "algorithm-4458b593b07f4ee1b06db455c9f09209",
"keywords": [
"leetcode",
"单词接龙 II"
],
"children": [],
"export": [
"solution.json"
],
"title": "单词接龙 II"
}
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/7.126-单词接龙 II/desc.html 0 → 100644
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<p>按字典 <code>wordList</code> 完成从单词 <code>beginWord</code> 到单词 <code>endWord</code> 转化，一个表示此过程的 <strong>转换序列</strong> 是形式上像 <code>beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub></code> 这样的单词序列，并满足：</p>
<div class="original__bRMd">
<div>
<ul>
<li>每对相邻的单词之间仅有单个字母不同。</li>
<li>转换过程中的每个单词 <code>s<sub>i</sub></code>（<code>1 <= i <= k</code>）必须是字典 <code>wordList</code> 中的单词。注意，<code>beginWord</code> 不必是字典 <code>wordList</code> 中的单词。</li>
<li><code>s<sub>k</sub> == endWord</code></li>
</ul>
<p>给你两个单词 <code>beginWord</code> 和 <code>endWord</code> ，以及一个字典 <code>wordList</code> 。请你找出并返回所有从 <code>beginWord</code> 到 <code>endWord</code> 的 <strong>最短转换序列</strong> ，如果不存在这样的转换序列，返回一个空列表。每个序列都应该以单词列表<em> </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code> 的形式返回。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
<strong>输出：</strong>[["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
<strong>解释：</strong>存在 2 种最短的转换序列：
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
<strong>输出：</strong>[]
<strong>解释：</strong>endWord "cog" 不在字典 wordList 中，所以不存在符合要求的转换序列。
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= beginWord.length <= 7</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 5000</code></li>
<li><code>wordList[i].length == beginWord.length</code></li>
<li><code>beginWord</code>、<code>endWord</code> 和 <code>wordList[i]</code> 由小写英文字母组成</li>
<li><code>beginWord != endWord</code></li>
<li><code>wordList</code> 中的所有单词 <strong>互不相同</strong></li>
</ul>
</div>
</div>
data/3.算法高阶/2.leetcode-哈希表/7.126-单词接龙 II/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
class Solution
{
public:
unordered_map<string, vector<string>> tree; // 构建图
vector<vector<string>> ans; // 存放最终结果
void dfs(vector<string> &cur, string curWord, string endWord)
{
if (curWord == endWord)
{
ans.push_back(cur);
return;
}
for (string s : tree[curWord])
{
cur.push_back(s);
dfs(cur, s, endWord);
cur.pop_back();
}
}
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
if (wordList.size() == 0 || find(wordList.begin(), wordList.end(), endWord) == wordList.end())
return {};
unordered_set<string> bfsFromBegin{beginWord}; // 自顶向下的BFS队列 !!!注意使用集合
unordered_set<string> bfsFromEnd{endWord}; // 自底向上的BFS队列 !!!注意使用集合
unordered_set<string> dirc(wordList.begin(), wordList.end()); // 初始化字典 记录未被访问过的字符串 !!!注意初始化方式
bool findFlag = false, reverseFlag = false; // findFlag两队列是否存在相同的元素 reverseflag时刻标记当前BFS遍历的方向（false为自顶向下，true为自底向上）
while (!bfsFromBegin.empty())
{
unordered_set<string> next; // 存放下一层需要遍历的节点（由于此处BFS的特殊性【一次性扩展所有节点】，所以不是直接添加到原先队列的末尾，而是借助next）
for (string s : bfsFromBegin) // 遍历过的节点从set中删除 避免成环
dirc.erase(s);
for (string s1 : bfsFromBegin)
{ // 扩展下一层的节点
for (int i = 0; i < s1.size(); ++i)
{
string s2 = s1; // s2记录由s1扩展而来字符串 !!!注意这条语句不要放错位置
for (char c = 'a'; c <= 'z'; ++c)
{
s2[i] = c;
if (dirc.count(s2) == 0)
continue;
if (bfsFromEnd.count(s2))
{
findFlag = true; // 找到双向BFS重合的字符串，BFS过程即可终止
}
else
{
next.insert(s2); // 将字符串加入扩展队列
}
reverseFlag ? tree[s2].push_back(s1) : tree[s1].push_back(s2); // 构建树 并始终保持方向从beginWord指向endWord
}
}
}
bfsFromBegin = next; // 更新队列
if (bfsFromBegin.size() > bfsFromEnd.size())
{
reverseFlag = !reverseFlag; // 取反
swap(bfsFromBegin, bfsFromEnd); // 交换BFS的队列 改变BFS的方向
}
if (findFlag)
break; // 双向BFS交汇 BFS过程终止
}
vector<string> cur = {beginWord};
dfs(cur, beginWord, endWord); // 遍历形成的树 得到起点到终点的路径
return ans;
}
};
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "84292d0d29094d59bd5be678c4c853e4"
}
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/7.126-单词接龙 II/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 单词接龙 II
<p>按字典 <code>wordList</code> 完成从单词 <code>beginWord</code> 到单词 <code>endWord</code> 转化，一个表示此过程的 <strong>转换序列</strong> 是形式上像 <code>beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub></code> 这样的单词序列，并满足：</p>
<div class="original__bRMd">
<div>
<ul>
<li>每对相邻的单词之间仅有单个字母不同。</li>
<li>转换过程中的每个单词 <code>s<sub>i</sub></code>（<code>1 <= i <= k</code>）必须是字典 <code>wordList</code> 中的单词。注意，<code>beginWord</code> 不必是字典 <code>wordList</code> 中的单词。</li>
<li><code>s<sub>k</sub> == endWord</code></li>
</ul>
<p>给你两个单词 <code>beginWord</code> 和 <code>endWord</code> ，以及一个字典 <code>wordList</code> 。请你找出并返回所有从 <code>beginWord</code> 到 <code>endWord</code> 的 <strong>最短转换序列</strong> ，如果不存在这样的转换序列，返回一个空列表。每个序列都应该以单词列表<em> </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code> 的形式返回。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
<strong>输出：</strong>[["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
<strong>解释：</strong>存在 2 种最短的转换序列：
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
<strong>输出：</strong>[]
<strong>解释：</strong>endWord "cog" 不在字典 wordList 中，所以不存在符合要求的转换序列。
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= beginWord.length <= 7</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 5000</code></li>
<li><code>wordList[i].length == beginWord.length</code></li>
<li><code>beginWord</code>、<code>endWord</code> 和 <code>wordList[i]</code> 由小写英文字母组成</li>
<li><code>beginWord != endWord</code></li>
<li><code>wordList</code> 中的所有单词 <strong>互不相同</strong></li>
</ul>
</div>
</div>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<vector<string>> res;
string beginWord = "hit", endWord = "cog";
vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};
res = sol.findLadders(beginWord, endWord, wordList);
for (auto i : res)
{
for (auto j : i)
cout << j << " ";
cout << endl;
}
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
unordered_map<string, vector<string>> tree;
vector<vector<string>> ans;
void dfs(vector<string> &cur, string curWord, string endWord)
{
if (curWord == endWord)
{
ans.push_back(cur);
return;
}
for (string s : tree[curWord])
{
cur.push_back(s);
dfs(cur, s, endWord);
cur.pop_back();
}
}
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
if (wordList.size() == 0 || find(wordList.begin(), wordList.end(), endWord) == wordList.end())
return {};
unordered_set<string> bfsFromBegin{beginWord};
unordered_set<string> bfsFromEnd{endWord};
unordered_set<string> dirc(wordList.begin(), wordList.end());
bool findFlag = false, reverseFlag = false;
while (!bfsFromBegin.empty())
{
unordered_set<string> next;
for (string s : bfsFromBegin)
dirc.erase(s);
for (string s1 : bfsFromBegin)
{
for (int i = 0; i < s1.size(); ++i)
{
string s2 = s1;
for (char c = 'a'; c <= 'z'; ++c)
{
s2[i] = c;
if (dirc.count(s2) == 0)
continue;
next.insert(s2);
}
reverseFlag ? tree[s2].push_back(s1) : tree[s1].push_back(s2);
}
}
bfsFromBegin = next;
if (bfsFromBegin.size() > bfsFromEnd.size())
{
reverseFlag = !reverseFlag;
swap(bfsFromBegin, bfsFromEnd);
}
if (findFlag)
break;
}
vector<string> cur = {beginWord};
dfs(cur, beginWord, endWord);
return ans;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
bool differ_one(string &s, string &t)
{
int n = 0;
for (int i = 0; i < s.size(); ++i)
if ((n += s[i] != t[i]) > 1)
return false;
return n == 1;
}
vector<vector<int>> fa;
vector<string> tmp;
vector<vector<string>> ans;
void dfs(int index, vector<string> &wordList)
{
if (fa[index].empty())
{
reverse(tmp.begin(), tmp.end());
ans.push_back(tmp);
reverse(tmp.begin(), tmp.end());
}
for (auto &x : fa[index])
{
tmp.push_back(wordList[x]);
dfs(x, wordList);
tmp.pop_back();
}
}
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
int b = -1, e = -1, x;
for (int i = 0; i < wordList.size(); ++i)
{
if (wordList[i] == beginWord)
b = i;
if (wordList[i] == endWord)
e = i;
}
if (e == -1)
return ans;
if (b == -1)
wordList.push_back(beginWord), b = wordList.size() - 1;
queue<int> que;
fa.assign(wordList.size(), vector<int>());
vector<int> index(wordList.size(), 0);
que.push(b), index[b] = 1;
while (!que.empty())
{
x = que.front(), que.pop();
if (index[e] && index[x] >= index[e])
break;
for (int i = 0; i < wordList.size(); ++i)
if ((index[i] == 0 || index[x] + 1 == index[i]) && differ_one(wordList[x], wordList[i]))
if (index[i] == 0)
index[i] = index[x] + 1, que.push(i), fa[i].push_back(x);
else
fa[i].push_back(x);
}
if (index[e] == 0)
return ans;
tmp.push_back(endWord);
dfs(e, wordList);
tmp.pop_back();
return ans;
}
};
```
### B
```cpp
class Solution
{
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> list(wordList.begin(), wordList.end());
if (!list.count(endWord))
return {};
unordered_map<string, unordered_set<string>> trace;
unordered_set<string> p;
unordered_set<string> q{beginWord};
while (q.size() && !trace.count(endWord))
{
for (auto w : q)
{
list.erase(w);
}
p.clear();
for (auto str : q)
{
for (int i = 0; i < str.size(); ++i)
{
string tmp = str;
for (char c = 'a'; c <= 'z'; ++c)
{
tmp[i] = c;
if (list.find(tmp) == list.end())
continue;
trace[tmp].insert(str);
p.insert(tmp);
}
}
}
q = p;
}
if (!trace.size())
return {};
vector<vector<string>> res;
dfs(res, {}, trace, endWord);
return res;
}
void dfs(vector<vector<string>> &res, vector<string> path, unordered_map<string, unordered_set<string>> &trace, string &str)
{
path.push_back(str);
if (trace.count(str) == 0)
{
reverse(path.begin(), path.end());
res.push_back(path);
return;
}
for (auto word : trace[str])
dfs(res, path, trace, word);
}
};
```
### C
```cpp
class Solution
{
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
vector<vector<string>> ans;
unordered_set<string> dict(wordList.begin(), wordList.end());
vector<string> p{beginWord};
queue<vector<string>> paths;
paths.push(p);
unordered_set<string> words;
int level = 1, minlength = INT_MAX;
while (!paths.empty())
{
vector<string> t = paths.front();
paths.pop();
if (t.size() > level)
{
for (string s : words)
dict.erase(s);
words.clear();
level = t.size();
if (level > minlength)
break;
}
string last = t.back();
for (int i = 0; i < last.size(); ++i)
{
string newlast = last;
for (char ch = 'a'; ch <= 'z'; ++ch)
{
newlast[i] = ch;
if (!dict.count(newlast))
continue;
words.insert(newlast);
vector<string> nextpath = t;
nextpath.push_back(newlast);
if (newlast == endWord)
{
ans.push_back(nextpath);
minlength = level;
}
else
{
paths.push(nextpath);
}
}
}
}
return ans;
}
};
```
data/3.算法高阶/2.leetcode-哈希表/8.127-单词接龙/config.json 0 → 100644
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{
"node_id": "algorithm-6df957f509fc40d1ae8a9736bd2401dc",
"keywords": [
"leetcode",
"单词接龙"
],
"children": [],
"export": [
"solution.json"
],
"title": "单词接龙"
}
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/8.127-单词接龙/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p>字典 <code>wordList</code> 中从单词 <code>beginWord</code><em> </em>和 <code>endWord</code> 的 <strong>转换序列 </strong>是一个按下述规格形成的序列：</p>
<ul>
<li>序列中第一个单词是 <code>beginWord</code> 。</li>
<li>序列中最后一个单词是 <code>endWord</code> 。</li>
<li>每次转换只能改变一个字母。</li>
<li>转换过程中的中间单词必须是字典 <code>wordList</code> 中的单词。</li>
</ul>
<p>给你两个单词<em> </em><code>beginWord</code><em> </em>和 <code>endWord</code> 和一个字典 <code>wordList</code> ，找到从 <code>beginWord</code> 到 <code>endWord</code> 的 <strong>最短转换序列</strong> 中的 <strong>单词数目</strong> 。如果不存在这样的转换序列，返回 0。</p>
 
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
<strong>输出：</strong>5
<strong>解释：</strong>一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
<strong>输出：</strong>0
<strong>解释：</strong>endWord "cog" 不在字典中，所以无法进行转换。</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= beginWord.length <= 10</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 5000</code></li>
<li><code>wordList[i].length == beginWord.length</code></li>
<li><code>beginWord</code>、<code>endWord</code> 和 <code>wordList[i]</code> 由小写英文字母组成</li>
<li><code>beginWord != endWord</code></li>
<li><code>wordList</code> 中的所有字符串 <strong>互不相同</strong></li>
</ul>
data/3.算法高阶/2.leetcode-哈希表/8.127-单词接龙/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
//Dijkstra算法的思路
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> set(wordList.begin(), wordList.end());
if (!set.count(endWord))
return 0;
queue<pair<string, int>> q; //<字符串，该字符串到beginword的距离>
q.push({beginWord, 1});
while (!q.empty())
{
string cur = q.front().first;
int step = q.front().second;
q.pop();
if (cur == endWord)
return step;
for (int i = 0; i < cur.size(); ++i)
{
char ch = cur[i];
//遍历字母来寻找距离为1（只用改变一个字母）的单词，也即可达的单词
for (char c = 'a'; c <= 'z'; ++c)
{
if (c == ch)
continue;
cur[i] = c; //替换单个字符
if (set.count(cur))
{
q.push({cur, 1 + step});
set.erase(cur); //该节点使用过了，需要进行删除
}
}
cur[i] = ch; //复原
}
}
return 0;
}
};
data/3.算法高阶/2.leetcode-哈希表/8.127-单词接龙/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "3cee91c0e09040b7b0c46b7f5ebb7cc5"
}
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/8.127-单词接龙/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 单词接龙
<p>字典 <code>wordList</code> 中从单词 <code>beginWord</code><em> </em>和 <code>endWord</code> 的 <strong>转换序列 </strong>是一个按下述规格形成的序列：</p>
<ul>
<li>序列中第一个单词是 <code>beginWord</code> 。</li>
<li>序列中最后一个单词是 <code>endWord</code> 。</li>
<li>每次转换只能改变一个字母。</li>
<li>转换过程中的中间单词必须是字典 <code>wordList</code> 中的单词。</li>
</ul>
<p>给你两个单词<em> </em><code>beginWord</code><em> </em>和 <code>endWord</code> 和一个字典 <code>wordList</code> ，找到从 <code>beginWord</code> 到 <code>endWord</code> 的 <strong>最短转换序列</strong> 中的 <strong>单词数目</strong> 。如果不存在这样的转换序列，返回 0。</p>
 
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
<strong>输出：</strong>5
<strong>解释：</strong>一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
<strong>输出：</strong>0
<strong>解释：</strong>endWord "cog" 不在字典中，所以无法进行转换。</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= beginWord.length <= 10</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 5000</code></li>
<li><code>wordList[i].length == beginWord.length</code></li>
<li><code>beginWord</code>、<code>endWord</code> 和 <code>wordList[i]</code> 由小写英文字母组成</li>
<li><code>beginWord != endWord</code></li>
<li><code>wordList</code> 中的所有字符串 <strong>互不相同</strong></li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
int res;
string beginWord = "hit", endWord = "cog";
vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};
res = sol.ladderLength(beginWord, endWord, wordList);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> word;
for (auto s : wordList)
{
word.insert(s);
}
if (word.find(endWord) == word.end())
{
return 0;
}
queue<string> que;
que.push(beginWord);
unordered_set<string> visited;
visited.insert(beginWord);
int num = 1;
while (!que.empty())
{
int m = que.size();
for (int i = 0; i < m; i++)
{
string str = que.front();
que.pop();
if (str == endWord)
{
return num;
}
for (int j = 0; j < str.size(); j++)
{
string ss = str;
for (int k = 0; k < 26; k++)
{
char c = k + 'a';
if (c != str[j])
{
ss[j] = c;
}
if (visited.find(ss) == visited.end())
{
visited.insert(ss);
que.push(ss);
}
}
}
}
num++;
}
return 0;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> wordSet(wordList.begin(), wordList.end());
if (!wordSet.count(endWord))
return 0;
unordered_map<string, int> pathCount{{{beginWord, 1}}};
queue<string> q{{beginWord}};
while (!q.empty())
{
string word = q.front();
q.pop();
for (int i = 0; i < word.size(); i++)
{
string newWord = word;
for (char c = 'a'; c <= 'z'; c++)
{
newWord[i] = c;
if (wordSet.count(newWord) && newWord == endWord)
return pathCount[word] + 1;
if (wordSet.count(newWord) && !pathCount.count(newWord))
{
pathCount[newWord] = pathCount[word] + 1;
q.push(newWord);
}
}
}
}
return 0;
}
};
```
### B
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
queue<string> q;
map<string, int> m1;
map<string, int> re;
int n = wordList.size();
for (int i = 0; i < n; i++)
m1[wordList[i]] = 1;
re[beginWord] = 1;
q.push(beginWord);
while ((!q.empty()) && m1.size())
{
string now = q.front();
q.pop();
int num = re[now];
int llen = now.size();
for (int i = 0; i < llen; i++)
{
string temp = now;
for (char c = 'a'; c <= 'z'; c++)
{
if (temp[i] == c)
continue;
else
temp[i] = c;
if (m1.find(temp) != m1.end())
{
if (temp == endWord)
return num + 1;
q.push(temp);
re[temp] = num + 1;
m1.erase(temp);
}
}
}
}
return 0;
}
};
```
### C
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> set(wordList.begin(), wordList.end());
if (!set.count(endWord))
return 0;
queue<pair<string, int>> q;
q.push({beginWord, 1});
while (!q.empty())
{
string cur = q.front().first;
int step = q.front().second;
q.pop();
if (cur == endWord)
return step;
for (int i = 0; i < cur.size(); ++i)
{
char ch = cur[i];
for (char c = 'a'; c <= 'z'; ++c)
{
if (c == ch)
continue;
cur[i] = c;
if (set.count(cur))
{
q.push({cur, 1 + step});
set.erase(cur);
}
}
cur[i] = ch;
}
}
return 0;
}
};
```
data/3.算法高阶/2.leetcode-哈希表/9.128-最长连续序列/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-78586cd84fd946a3ab3df877877a7a70",
"keywords": [
"leetcode",
"最长连续序列"
],
"children": [],
"export": [
"solution.json"
],
"title": "最长连续序列"
}
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/9.128-最长连续序列/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p>给定一个未排序的整数数组 <code>nums</code> ，找出数字连续的最长序列（不要求序列元素在原数组中连续）的长度。</p>
<p>请你设计并实现时间复杂度为 <code>O(n)</code><em> </em>的算法解决此问题。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>nums = [100,4,200,1,3,2]
<strong>输出：</strong>4
<strong>解释：</strong>最长数字连续序列是 <code>[1, 2, 3, 4]。它的长度为 4。</code></pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>nums = [0,3,7,2,5,8,4,6,0,1]
<strong>输出：</strong>9
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>0 <= nums.length <= 10<sup>5</sup></code></li>
<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>
</ul>
data/3.算法高阶/2.leetcode-哈希表/9.128-最长连续序列/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
#include <vector>
#include <unordered_map>
using std::unordered_map;
using std::vector;
class Solution
{
public:
int longestConsecutive(vector<int> &num)
{
unordered_map<int, int> map;
int max{0};
for (auto i : num)
if (!map[i])
{
map[i] = map[i + 1] + map[i - 1] + 1;
map[i - map[i - 1]] = map[i + map[i + 1]] = map[i];
max = std::max(max, map[i]);
}
return max;
}
};
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/9.128-最长连续序列/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "dce3e3515a4b405e8aec92b075fac20f"
}
\ No newline at end of file
data/3.算法高阶/2.leetcode-哈希表/9.128-最长连续序列/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 最长连续序列
<p>给定一个未排序的整数数组 <code>nums</code> ，找出数字连续的最长序列（不要求序列元素在原数组中连续）的长度。</p>
<p>请你设计并实现时间复杂度为 <code>O(n)</code><em> </em>的算法解决此问题。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>nums = [100,4,200,1,3,2]
<strong>输出：</strong>4
<strong>解释：</strong>最长数字连续序列是 <code>[1, 2, 3, 4]。它的长度为 4。</code></pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>nums = [0,3,7,2,5,8,4,6,0,1]
<strong>输出：</strong>9
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>0 <= nums.length <= 10<sup>5</sup></code></li>
<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
int res;
vector<int> vec = {0, 3, 7, 2, 5, 8, 4, 6, 0, 1};
res = sol.longestSequence(vec);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
int longestConsecutive(vector<int> &nums)
{
unordered_set<int> hash;
for (const int &num : nums)
{
hash.insert(num);
}
int ans = 0;
while (!hash.empty())
{
int cur = *(hash.begin());
hash.erase(cur);
int next = cur + 1, prev = cur - 1;
while (hash.count(next))
{
hash.erase(next++);
}
while (hash.count(prev))
{
hash.erase(prev--);
}
ans = max(ans, next - prev);
}
return ans;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int longestConsecutive(vector<int> &nums)
{
unordered_map<int, int> mp;
int l = 0, r = 0, res = 0, len = 0;
for (int num : nums)
{
if (!mp[num])
{
l = mp[num - 1];
r = mp[num + 1];
len = l + r + 1;
res = max(res, len);
mp[num] = len;
mp[num - l] = len;
mp[num + r] = len;
}
}
return res;
}
};
```
### B
```cpp
class Solution
{
struct DisJointSet
{
vector<int> _id;
vector<int> _size;
int max_size;
int _count;
DisJointSet(int Num)
{
for (int i = 0; i < Num; i++)
{
_id.emplace_back(i);
_size.emplace_back(1);
}
_count = Num;
max_size = 1;
}
int find_(int p)
{
while (p != _id[p])
{
_id[p] = _id[_id[p]];
p = _id[p];
}
return p;
}
void _union(int p, int q)
{
int i = find_(p);
int j = find_(q);
if (i == j)
return;
if (_size[i] > _size[j])
{
_id[j] = i;
_size[i] += _size[j];
max_size = max(max_size, _size[i]);
}
else
{
_id[i] = j;
_size[j] += _size[i];
max_size = max(max_size, _size[j]);
}
_count--;
}
};
public:
int longestConsecutive(vector<int> &nums)
{
if (nums.size() == 0)
return 0;
DisJointSet disJointSet(nums.size());
unordered_set<int> nums_set;
unordered_map<int, int> nums_disJointSetID_map;
for (int i = 0; i < nums.size(); i++)
{
if (nums_set.find(nums[i]) != nums_set.end())
continue;
nums_set.insert(nums[i]);
nums_disJointSetID_map[nums[i]] = i;
if (nums_set.find(nums[i] - 1) != nums_set.end())
{
disJointSet._union(nums_disJointSetID_map[nums[i]], nums_disJointSetID_map[nums[i] - 1]);
}
if (nums_set.find(nums[i] + 1) != nums_set.end())
{
disJointSet._union(nums_disJointSetID_map[nums[i]], nums_disJointSetID_map[nums[i] + 1]);
}
}
return disJointSet.max_size;
}
};
```
### C
```cpp
class Solution
{
public:
int longestConsecutive(vector<int> &vec)
{
set<int> s;
for (int elem : vec)
s.insert(elem);
int longest = 0;
int cnt = 1;
for (auto e : s)
{
if (s.find(e + 1) != s.end())
{
cnt++;
}
else
{
cnt = 1;
}
longest = max(longest, cnt);
}
return longest;
}
};
```
data/3.算法高阶/3.leetcode-图与搜索/10.130-被围绕的区域/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-d221fe9f0a114efdbd615b8cba03e7fc",
"keywords": [
"leetcode",
"被围绕的区域"
],
"children": [],
"export": [
"solution.json"
],
"title": "被围绕的区域"
}
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/10.130-被围绕的区域/desc.html 0 → 100644
浏览文件 @ 66b8b16b
给你一个 <code>m x n</code> 的矩阵 <code>board</code> ，由若干字符 <code>'X'</code> 和 <code>'O'</code> ，找到所有被 <code>'X'</code> 围绕的区域，并将这些区域里所有的 <code>'O'</code> 用 <code>'X'</code> 填充。
<div class="original__bRMd">
<div>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/xogrid.jpg" style="width: 550px; height: 237px;" />
<pre>
<strong>输入：</strong>board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
<strong>输出：</strong>[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
<strong>解释：</strong>被围绕的区间不会存在于边界上，换句话说，任何边界上的 <code>'O'</code> 都不会被填充为 <code>'X'</code>。 任何不在边界上，或不与边界上的 <code>'O'</code> 相连的 <code>'O'</code> 最终都会被填充为 <code>'X'</code>。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>board = [["X"]]
<strong>输出：</strong>[["X"]]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>m == board.length</code></li>
<li><code>n == board[i].length</code></li>
<li><code>1 <= m, n <= 200</code></li>
<li><code>board[i][j]</code> 为 <code>'X'</code> 或 <code>'O'</code></li>
</ul>
</div>
</div>
data/3.算法高阶/3.leetcode-图与搜索/10.130-被围绕的区域/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
class Solution
{
public:
int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0}, n, m;
void dfs(int row, int col, vector<vector<char>> &board)
{
board[row][col] = 'o';
for (int i = 0; i < 4; ++i)
if (row + dx[i] >= 0 && row + dx[i] < n && col + dy[i] >= 0 && col + dy[i] < m && board[row + dx[i]][col + dy[i]] == 'O')
dfs(row + dx[i], col + dy[i], board);
}
void solve(vector<vector<char>> &board)
{
n = board.size(), m = n ? board[0].size() : 0;
if (!(m & n))
return;
for (int i = 0; i < n; ++i)
{
if (board[i][0] == 'O')
dfs(i, 0, board);
if (board[i][m - 1] == 'O')
dfs(i, m - 1, board);
}
for (int j = 0; j < m; ++j)
{
if (board[0][j] == 'O')
dfs(0, j, board);
if (board[n - 1][j] == 'O')
dfs(n - 1, j, board);
}
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j)
if (board[i][j] == 'O')
board[i][j] = 'X';
else if (board[i][j] == 'o')
board[i][j] = 'O';
}
};
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/10.130-被围绕的区域/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "a2608cf9adcf469d86ac1874392f6288"
}
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/10.130-被围绕的区域/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 被围绕的区域
给你一个 <code>m x n</code> 的矩阵 <code>board</code> ，由若干字符 <code>'X'</code> 和 <code>'O'</code> ，找到所有被 <code>'X'</code> 围绕的区域，并将这些区域里所有的 <code>'O'</code> 用 <code>'X'</code> 填充。
<div class="original__bRMd">
<div>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/xogrid.jpg" style="width: 550px; height: 237px;" />
<pre>
<strong>输入：</strong>board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
<strong>输出：</strong>[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
<strong>解释：</strong>被围绕的区间不会存在于边界上，换句话说，任何边界上的 <code>'O'</code> 都不会被填充为 <code>'X'</code>。 任何不在边界上，或不与边界上的 <code>'O'</code> 相连的 <code>'O'</code> 最终都会被填充为 <code>'X'</code>。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>board = [["X"]]
<strong>输出：</strong>[["X"]]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>m == board.length</code></li>
<li><code>n == board[i].length</code></li>
<li><code>1 <= m, n <= 200</code></li>
<li><code>board[i][j]</code> 为 <code>'X'</code> 或 <code>'O'</code></li>
</ul>
</div>
</div>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<vector<char>> board = {{'X', 'X', 'X', 'X'}, {'X', 'O', 'O', 'X'}, {'X', 'X', 'O', 'X'}, {'X', 'O', 'X', 'X'}};
sol.solve(board);
for (auto i : board)
{
for (auto j : i)
cout << j << " ";
cout << endl;
}
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
vector<vector<bool>> st;
int m, n;
void bfs(vector<vector<char>> &board, int a, int b)
{
int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
st[a][b] = true;
queue<pair<int, int>> q;
q.push(make_pair(a, b));
while (q.size())
{
pair<int, int> t = q.front();
q.pop();
for (int i = 0; i < 4; i++)
{
int x = dx[i] + t.first, y = dy[i] + t.second;
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O' && !st[x][y])
{
q.push(make_pair(x, y));
st[x][y] = true;
}
}
}
}
void solve(vector<vector<char>> &board)
{
if (!board.size())
return;
m = board.size(), n = board[0].size();
st = vector<vector<bool>>(m, vector<bool>(n, false));
for (int i = 0; i < m; i++)
{
if (board[i][0] == 'O')
bfs(board, i, 0);
if (board[i][n] == 'O')
bfs(board, i, n - 1);
}
for (int i = 0; i < n; i++)
{
if (board[0][i] == 'O')
bfs(board, 0, i);
if (board[m][i] == 'O')
bfs(board, m - 1, i);
}
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (!st[i][j])
board[i][j] = 'X';
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int n, m;
void dfs(vector<vector<char>> &board, int x, int y)
{
if (x < 0 || y < 0 || x >= m || y >= n || board[x][y] != 'O')
return;
board[x][y] = 'A';
dfs(board, x + 1, y);
dfs(board, x, y + 1);
dfs(board, x - 1, y);
dfs(board, x, y - 1);
}
void solve(vector<vector<char>> &board)
{
m = board.size();
if (m == 0)
return;
n = board[0].size();
for (int i = 0; i < m; i++)
{
dfs(board, i, 0);
dfs(board, i, n - 1);
}
for (int i = 1; i < n - 1; i++)
{
dfs(board, 0, i);
dfs(board, m - 1, i);
}
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
if (board[i][j] == 'A')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
}
};
```
### B
```cpp
class Solution
{
public:
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
queue<pair<int, int>> s;
void solve(vector<vector<char>> &board)
{
int n = board.size();
if (n <= 2)
return;
int m = board[0].size();
if (m <= 2)
return;
for (int i = 0; i < m; i++)
{
if (board[0][i] == 'O')
s.push({0, i});
if (board[n - 1][i] == 'O')
s.push({n - 1, i});
}
for (int i = 1; i < n - 1; i++)
{
if (board[i][0] == 'O')
s.push({i, 0});
if (board[i][m - 1] == 'O')
s.push({i, m - 1});
}
while (!s.empty())
{
int x = s.front().first, y = s.front().second;
board[x][y] = '#';
s.pop();
for (int i = 0; i < 4; i++)
{
int mx = x + dx[i], my = y + dy[i];
if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O')
{
continue;
}
s.push({mx, my});
}
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (board[i][j] == '#')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
}
};
```
### C
```cpp
class Solution
{
public:
void dfs(vector<vector<char>> &board, int row, int col)
{
if (row < 0 || row >= board.size() || col < 0 || col >= board[0].size())
return;
if (board[row][col] != 'O')
return;
board[row][col] = '+';
dfs(board, row + 1, col);
dfs(board, row, col + 1);
dfs(board, row - 1, col);
dfs(board, row, col - 1);
}
void solve(vector<vector<char>> &board)
{
if (board.size() <= 1 || board[0].size() <= 2)
return;
for (int i = 0; i < board.size(); i++)
{
if (board[i][0] == 'O')
dfs(board, i, 0);
if (board[i][board[0].size() - 1] == 'O')
dfs(board, i, board[0].size() - 1);
}
for (int i = 0; i < board[0].size(); i++)
{
if (board[0][i] == 'O')
dfs(board, 0, i);
if (board[board.size() - 1][i] == 'O')
dfs(board, board.size() - 1, i);
}
for (int i = 0; i < board.size(); i++)
{
for (int j = 0; j < board[0].size(); j++)
{
if (board[i][j] == '+')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
}
};
```
data/3.算法高阶/3.leetcode-图与搜索/8.112-路径总和/config.json 0 → 100644
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{
"node_id": "algorithm-15e7a6d4c7da4ed682782b7b4b9761d2",
"keywords": [
"leetcode",
"路径总和"
],
"children": [],
"export": [
"solution.json"
],
"title": "路径总和"
}
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/8.112-路径总和/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p>给你二叉树的根节点 <code>root</code> 和一个表示目标和的整数 <code>targetSum</code> ，判断该树中是否存在 <strong>根节点到叶子节点</strong> 的路径，这条路径上所有节点值相加等于目标和 <code>targetSum</code> 。</p>
<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
<strong>输出：</strong>true
</pre>
<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>false
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>root = [1,2], targetSum = 0
<strong>输出：</strong>false
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点的数目在范围 <code>[0, 5000]</code> 内</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>
data/3.算法高阶/3.leetcode-图与搜索/8.112-路径总和/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
#include <cstddef>
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
bool hasPathSum(TreeNode *root, int sum)
{
if (!root)
return false;
if (!root->left and !root->right and sum - root->val == 0)
return true;
return hasPathSum(root->left, sum - root->val) or hasPathSum(root->right, sum - root->val);
}
};
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/8.112-路径总和/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "544c6d3b7e0d4116b27d5fdb8691a522"
}
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/8.112-路径总和/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 路径总和
<p>给你二叉树的根节点 <code>root</code> 和一个表示目标和的整数 <code>targetSum</code> ，判断该树中是否存在 <strong>根节点到叶子节点</strong> 的路径，这条路径上所有节点值相加等于目标和 <code>targetSum</code> 。</p>
<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
<strong>输出：</strong>true
</pre>
<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>false
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>root = [1,2], targetSum = 0
<strong>输出：</strong>false
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点的数目在范围 <code>[0, 5000]</code> 内</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
```
## 答案
```cpp
其他三个都是错的
```
## 选项
### A
```cpp
class Solution
{
public:
bool hasPathSum(TreeNode *root, int sum)
{
if (root == NULL)
{
return false;
}
vector<TreeNode *> node_stack;
vector<int> sum_stack;
node_stack.push_back(root);
sum_stack.push_back(sum - root->val);
TreeNode *node;
int curr_sum;
while (!node_stack.empty())
{
node = node_stack.back();
node_stack.pop_back();
curr_sum = sum_stack.back();
sum_stack.pop_back();
if ((node->right == NULL) && (node->left == NULL) && (curr_sum == 0))
{
return true;
}
if (node->right != NULL)
{
node_stack.push_back(node->right);
sum_stack.push_back(curr_sum - node->right->val);
}
if (node->left != NULL)
{
node_stack.push_back(node->left);
sum_stack.push_back(curr_sum - node->left->val);
}
}
return false;
}
};
```
### B
```cpp
class Solution
{
public:
bool hasPathSum(TreeNode *root, int sum)
{
if (root == NULL)
return false;
stack<pair<TreeNode *, int>> s;
s.push(make_pair(root, sum));
TreeNode *node = root;
int currSum;
while (!s.empty())
{
node = s.top().first;
currSum = s.top().second;
s.pop();
if (node->left == NULL && node->right == NULL && node->val == currSum)
return true;
if (node->right)
s.push(make_pair(node->right, currSum - node->val));
if (node->left)
s.push(make_pair(node->left, currSum - node->val));
}
return false;
}
};
```
### C
```cpp
class Solution
{
public:
bool hasPathSum(TreeNode *root, int sum)
{
if (!root)
return false;
if (!root->left && !root->right && root->val == sum)
return true;
return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}
};
```
data/3.算法高阶/3.leetcode-图与搜索/9.113-路径总和 II/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-32f257cc9a114a4cb092581b3a017d74",
"keywords": [
"leetcode",
"路径总和 II"
],
"children": [],
"export": [
"solution.json"
],
"title": "路径总和 II"
}
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/9.113-路径总和 II/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p>给你二叉树的根节点 <code>root</code> 和一个整数目标和 <code>targetSum</code> ，找出所有 <strong>从根节点到叶子节点</strong> 路径总和等于给定目标和的路径。</p>
<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>
<div class="original__bRMd">
<div>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsumii1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>输出：</strong>[[5,4,11,2],[5,8,4,5]]
</pre>
<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" style="width: 212px; height: 181px;" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>[]
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>root = [1,2], targetSum = 0
<strong>输出：</strong>[]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点总数在范围 <code>[0, 5000]</code> 内</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>
</div>
</div>
data/3.算法高阶/3.leetcode-图与搜索/9.113-路径总和 II/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
#include <vector>
#include <cstddef>
#include <stack>
#include <algorithm>
using std::vector;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode *root, int sum)
{
vector<vector<int>> ret;
vector<int> path;
pathSum(root, sum, path, ret);
return ret;
}
void pathSum(TreeNode *root, int sum, vector<int> &path, vector<vector<int>> &ret)
{
if (!root)
return;
path.push_back(root->val);
if (!root->left && !root->right && root->val == sum)
ret.push_back(path);
if (root->left)
pathSum(root->left, sum - root->val, path, ret);
if (root->right)
pathSum(root->right, sum - root->val, path, ret);
path.pop_back();
}
};
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/9.113-路径总和 II/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "cebd88413a964aeca882afdf79cb545a"
}
\ No newline at end of file
data/3.算法高阶/3.leetcode-图与搜索/9.113-路径总和 II/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 路径总和 II
<p>给你二叉树的根节点 <code>root</code> 和一个整数目标和 <code>targetSum</code> ，找出所有 <strong>从根节点到叶子节点</strong> 路径总和等于给定目标和的路径。</p>
<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>
<div class="original__bRMd">
<div>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsumii1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>输出：</strong>[[5,4,11,2],[5,8,4,5]]
</pre>
<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" style="width: 212px; height: 181px;" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>[]
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>root = [1,2], targetSum = 0
<strong>输出：</strong>[]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点总数在范围 <code>[0, 5000]</code> 内</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>
</div>
</div>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
```
## 答案
```cpp
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode *root, int sum)
{
int pos = 0;
vector<int> path;
vector<vector<int>> result;
dfs(root, sum, pos, path, result);
return result;
}
private:
void dfs(TreeNode *root, int sum, int &pos, vector<int> &path, vector<vector<int>> &result)
{
if (root == NULL)
return;
pos += root->val;
path.push_back(root->val);
if (!root->left && !root->right && pos == sum)
result.push_back(path);
dfs(root->left, sum, pos, path, result);
dfs(root->right, sum, pos, path, result);
path.pop_back();
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
void find(TreeNode *node, vector<vector<int>> &result, vector<int> &path, int sum)
{
path.push_back(node->val);
if (node->left == NULL && node->right == NULL && sum == node->val)
{
result.push_back(path);
return;
}
if (node->left != NULL)
{
find(node->left, result, path, sum - node->val);
path.pop_back();
}
if (node->right != NULL)
{
find(node->right, result, path, sum - node->val);
path.pop_back();
}
}
vector<vector<int>> pathSum(TreeNode *root, int sum)
{
vector<vector<int>> result;
vector<int> path;
if (root == NULL)
{
return result;
}
find(root, result, path, sum);
return result;
}
};
```
### B
```cpp
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode *root, int sum)
{
vector<vector<int>> res;
if (!root)
return res;
vector<int> temp;
dfs(root, res, temp, sum);
return res;
}
void dfs(TreeNode *root, vector<vector<int>> &res, vector<int> temp, int sum)
{
temp.push_back(root->val);
if (!root->left && !root->right && 0 == sum - root->val)
res.push_back(temp);
if (root->left)
dfs(root->left, res, temp, sum - root->val);
if (root->right)
dfs(root->right, res, temp, sum - root->val);
}
};
```
### C
```cpp
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode *root, int targetSum)
{
DFS(root, targetSum);
return result;
}
void DFS(TreeNode *root, int targetSum)
{
if (root == nullptr)
return;
tmp_path.emplace_back(root->val);
targetSum -= root->val;
if (root->left == nullptr && root->right == nullptr && targetSum == 0)
{
result.emplace_back(tmp_path);
}
DFS(root->left, targetSum);
DFS(root->right, targetSum);
tmp_path.pop_back();
}
private:
vector<vector<int>> result;
vector<int> tmp_path;
};
```
data/3.算法高阶/5.leetcode-设计/6.173-二叉搜索树迭代器/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-bfd16b54a79348af9077cc46f46e5169",
"keywords": [
"leetcode",
"二叉搜索树迭代器"
],
"children": [],
"export": [
"solution.json"
],
"title": "二叉搜索树迭代器"
}
\ No newline at end of file
data/3.算法高阶/5.leetcode-设计/6.173-二叉搜索树迭代器/desc.html 0 → 100644
浏览文件 @ 66b8b16b
实现一个二叉搜索树迭代器类<code>BSTIterator</code> ，表示一个按中序遍历二叉搜索树（BST）的迭代器：
<div class="original__bRMd">
<div>
<ul>
<li><code>BSTIterator(TreeNode root)</code> 初始化 <code>BSTIterator</code> 类的一个对象。BST 的根节点 <code>root</code> 会作为构造函数的一部分给出。指针应初始化为一个不存在于 BST 中的数字，且该数字小于 BST 中的任何元素。</li>
<li><code>boolean hasNext()</code> 如果向指针右侧遍历存在数字，则返回 <code>true</code> ；否则返回 <code>false</code> 。</li>
<li><code>int next()</code>将指针向右移动，然后返回指针处的数字。</li>
</ul>
<p>注意，指针初始化为一个不存在于 BST 中的数字，所以对 <code>next()</code> 的首次调用将返回 BST 中的最小元素。</p>
</div>
</div>
<p>你可以假设 <code>next()</code> 调用总是有效的，也就是说，当调用 <code>next()</code> 时，BST 的中序遍历中至少存在一个下一个数字。</p>
<p> </p>
<p><strong>示例：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png" style="width: 189px; height: 178px;" />
<pre>
<strong>输入</strong>
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
<strong>输出</strong>
[null, 3, 7, true, 9, true, 15, true, 20, false]
<strong>解释</strong>
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // 返回 3
bSTIterator.next(); // 返回 7
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 9
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 15
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 20
bSTIterator.hasNext(); // 返回 False
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点的数目在范围 <code>[1, 10<sup>5</sup>]</code> 内</li>
<li><code>0 <= Node.val <= 10<sup>6</sup></code></li>
<li>最多调用 <code>10<sup>5</sup></code> 次 <code>hasNext</code> 和 <code>next</code> 操作</li>
</ul>
<p> </p>
<p><strong>进阶：</strong></p>
<ul>
<li>你可以设计一个满足下述条件的解决方案吗？<code>next()</code> 和 <code>hasNext()</code> 操作均摊时间复杂度为 <code>O(1)</code> ，并使用 <code>O(h)</code> 内存。其中 <code>h</code> 是树的高度。</li>
</ul>
data/3.算法高阶/5.leetcode-设计/6.173-二叉搜索树迭代器/solution.cpp 0 → 100644
浏览文件 @ 66b8b16b
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class BSTIterator
{
public:
BSTIterator(TreeNode *root)
{
TreeNode *cur = root;
while (cur)
{
st.push(cur);
cur = cur->left;
}
}
/** @return whether we have a next smallest number */
bool hasNext()
{
return !st.empty();
}
/** @return the next smallest number */
int next()
{
TreeNode *next = st.top();
TreeNode *cur = next->right;
st.pop();
while (cur)
{
st.push(cur);
cur = cur->left;
}
return next->val;
}
private:
stack<TreeNode *> st;
};
data/3.算法高阶/5.leetcode-设计/6.173-二叉搜索树迭代器/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "860327443b0d4f12a8fb51c7d674c3e9"
}
\ No newline at end of file
data/3.算法高阶/5.leetcode-设计/6.173-二叉搜索树迭代器/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 二叉搜索树迭代器
实现一个二叉搜索树迭代器类<code>BSTIterator</code> ，表示一个按中序遍历二叉搜索树（BST）的迭代器：
<div class="original__bRMd">
<div>
<ul>
<li><code>BSTIterator(TreeNode root)</code> 初始化 <code>BSTIterator</code> 类的一个对象。BST 的根节点 <code>root</code> 会作为构造函数的一部分给出。指针应初始化为一个不存在于 BST 中的数字，且该数字小于 BST 中的任何元素。</li>
<li><code>boolean hasNext()</code> 如果向指针右侧遍历存在数字，则返回 <code>true</code> ；否则返回 <code>false</code> 。</li>
<li><code>int next()</code>将指针向右移动，然后返回指针处的数字。</li>
</ul>
<p>注意，指针初始化为一个不存在于 BST 中的数字，所以对 <code>next()</code> 的首次调用将返回 BST 中的最小元素。</p>
</div>
</div>
<p>你可以假设 <code>next()</code> 调用总是有效的，也就是说，当调用 <code>next()</code> 时，BST 的中序遍历中至少存在一个下一个数字。</p>
<p> </p>
<p><strong>示例：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png" style="width: 189px; height: 178px;" />
<pre>
<strong>输入</strong>
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
<strong>输出</strong>
[null, 3, 7, true, 9, true, 15, true, 20, false]
<strong>解释</strong>
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // 返回 3
bSTIterator.next(); // 返回 7
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 9
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 15
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 20
bSTIterator.hasNext(); // 返回 False
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点的数目在范围 <code>[1, 10<sup>5</sup>]</code> 内</li>
<li><code>0 <= Node.val <= 10<sup>6</sup></code></li>
<li>最多调用 <code>10<sup>5</sup></code> 次 <code>hasNext</code> 和 <code>next</code> 操作</li>
</ul>
<p> </p>
<p><strong>进阶：</strong></p>
<ul>
<li>你可以设计一个满足下述条件的解决方案吗？<code>next()</code> 和 <code>hasNext()</code> 操作均摊时间复杂度为 <code>O(1)</code> ，并使用 <code>O(h)</code> 内存。其中 <code>h</code> 是树的高度。</li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
/**
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator* obj = new BSTIterator(root);
* int param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/
```
## 答案
```cpp
class BSTIterator
{
private:
void traverse(TreeNode *root, vector<int> &res)
{
if (root == nullptr)
{
return;
}
traverse(root->right, res);
traverse(root->left, res);
res.push_back(root->val);
}
vector<int> inorder(TreeNode *root)
{
vector<int> res;
traverse(root, res);
return res;
}
vector<int> arr;
int idx;
public:
BSTIterator(TreeNode *root) : idx(0), arr(inorder(root))
{
}
int next()
{
return arr[idx++];
}
bool hasNext()
{
return idx < arr.size();
}
};
```
## 选项
### A
```cpp
class BSTIterator
{
private:
void inorder(TreeNode *root, vector<int> &vec)
{
if (!root)
return;
inorder(root->left, vec);
vec.emplace_back(root->val);
inorder(root->right, vec);
}
vector<int> inorderTraversal(TreeNode *root)
{
vector<int> ret;
inorder(root, ret);
return ret;
}
vector<int> buf;
size_t idx;
public:
BSTIterator(TreeNode *root)
{
idx = 0;
buf = inorderTraversal(root);
}
int next()
{
return buf[idx++];
}
bool hasNext()
{
return idx < buf.size();
}
};
```
### B
```cpp
class BSTIterator
{
public:
stack<TreeNode *> treeMin;
BSTIterator(TreeNode *root)
{
TreeNode *t = root;
while (t)
{
treeMin.push(t);
t = t->left;
}
}
/** @return the next smallest number */
int next()
{
TreeNode *tmp = treeMin.top();
int res = tmp->val;
treeMin.pop();
tmp = tmp->right;
while (tmp)
{
treeMin.push(tmp);
tmp = tmp->left;
}
return res;
}
/** @return whether we have a next smallest number */
bool hasNext()
{
if (treeMin.empty())
return false;
else
return true;
}
};
```
### C
```cpp
class BSTIterator
{
public:
queue<int> q;
BSTIterator(TreeNode *root)
{
inorder(root, q);
}
void inorder(TreeNode *root, queue<int> &q)
{
if (root == NULL)
return;
inorder(root->left, q);
q.push(root->val);
inorder(root->right, q);
}
/** @return the next smallest number */
int next()
{
int tmp = q.front();
q.pop();
return tmp;
}
/** @return whether we have a next smallest number */
bool hasNext()
{
if (q.empty())
return false;
else
return true;
}
};
```
data/3.算法高阶/5.leetcode-设计/7.208-实现 Trie (前缀树)/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-14d4d8e793b84c1cb3efe42497cfb78c",
"keywords": [
"leetcode",
"实现 Trie (前缀树)"
],
"children": [],
"export": [
"solution.json"
],
"title": "实现 Trie (前缀树)"
}
\ No newline at end of file
data/3.算法高阶/5.leetcode-设计/7.208-实现 Trie (前缀树)/desc.html 0 → 100644
浏览文件 @ 66b8b16b
<p><strong><a href="https://baike.baidu.com/item/字典树/9825209?fr=aladdin" target="_blank">Trie</a></strong>（发音类似 "try"）或者说 <strong>前缀树</strong> 是一种树形数据结构，用于高效地存储和检索字符串数据集中的键。这一数据结构有相当多的应用情景，例如自动补完和拼写检查。</p>
<p>请你实现 Trie 类：</p>
<ul>
<li><code>Trie()</code> 初始化前缀树对象。</li>
<li><code>void insert(String word)</code> 向前缀树中插入字符串 <code>word</code> 。</li>
<li><code>boolean search(String word)</code> 如果字符串 <code>word</code> 在前缀树中，返回 <code>true</code>（即，在检索之前已经插入）；否则，返回 <code>false</code> 。</li>
<li><code>boolean startsWith(String prefix)</code> 如果之前已经插入的字符串 <code>word</code> 的前缀之一为 <code>prefix</code> ，返回 <code>true</code> ；否则，返回 <code>false</code> 。</li>
</ul>
<p> </p>
<p><strong>示例：</strong></p>
<pre>
<strong>输入</strong>
["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
[[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
<strong>输出</strong>
[null, null, true, false, true, null, true]
<strong>解释</strong>
Trie trie = new Trie();
trie.insert("apple");
trie.search("apple"); // 返回 True
trie.search("app"); // 返回 False
trie.startsWith("app"); // 返回 True
trie.insert("app");
trie.search("app"); // 返回 True
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= word.length, prefix.length <= 2000</code></li>
<li><code>word</code> 和 <code>prefix</code> 仅由小写英文字母组成</li>
<li><code>insert</code>、<code>search</code> 和 <code>startsWith</code> 调用次数 <strong>总计</strong> 不超过 <code>3 * 10<sup>4</sup></code> 次</li>
</ul>
data/3.算法高阶/5.leetcode-设计/7.208-实现 Trie (前缀树)/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fc412fd05a0244d0a8cd4d86cf9e775b"
}
\ No newline at end of file
data/3.算法高阶/5.leetcode-设计/7.208-实现 Trie (前缀树)/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# 实现 Trie (前缀树)
<p><strong><a href="https://baike.baidu.com/item/字典树/9825209?fr=aladdin" target="_blank">Trie</a></strong>（发音类似 "try"）或者说 <strong>前缀树</strong> 是一种树形数据结构，用于高效地存储和检索字符串数据集中的键。这一数据结构有相当多的应用情景，例如自动补完和拼写检查。</p>
<p>请你实现 Trie 类：</p>
<ul>
<li><code>Trie()</code> 初始化前缀树对象。</li>
<li><code>void insert(String word)</code> 向前缀树中插入字符串 <code>word</code> 。</li>
<li><code>boolean search(String word)</code> 如果字符串 <code>word</code> 在前缀树中，返回 <code>true</code>（即，在检索之前已经插入）；否则，返回 <code>false</code> 。</li>
<li><code>boolean startsWith(String prefix)</code> 如果之前已经插入的字符串 <code>word</code> 的前缀之一为 <code>prefix</code> ，返回 <code>true</code> ；否则，返回 <code>false</code> 。</li>
</ul>
<p> </p>
<p><strong>示例：</strong></p>
<pre>
<strong>输入</strong>
["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
[[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
<strong>输出</strong>
[null, null, true, false, true, null, true]
<strong>解释</strong>
Trie trie = new Trie();
trie.insert("apple");
trie.search("apple"); // 返回 True
trie.search("app"); // 返回 False
trie.startsWith("app"); // 返回 True
trie.insert("app");
trie.search("app"); // 返回 True
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= word.length, prefix.length <= 2000</code></li>
<li><code>word</code> 和 <code>prefix</code> 仅由小写英文字母组成</li>
<li><code>insert</code>、<code>search</code> 和 <code>startsWith</code> 调用次数 <strong>总计</strong> 不超过 <code>3 * 10<sup>4</sup></code> 次</li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
```
## 答案
```cpp
class TrieNode
{
public:
bool end_with;
TrieNode *next[26];
TrieNode()
{
end_with = false;
memset(next, 0, sizeof(next));
}
};
class Trie
{
TrieNode root;
public:
/** Initialize your data structure here. */
Trie() {}
/** Inserts a word into the trie. */
void insert(string word)
{
TrieNode *node = &root;
for (int i = 0; i < word.size(); i++)
{
int c = word[i] - 'a';
node->next[c] = new TrieNode();
node = node->next[c];
}
node->end_with = true;
}
/** Returns if the word is in the trie. */
bool search(string word)
{
TrieNode *node = &root;
for (int i = 0; i < word.size(); i++)
{
if (node->next[word[i] - 'a'] != NULL)
node = node->next[word[i] - 'a'];
else
return false;
}
return (node->end_with) ? true : false;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
bool startsWith(string prefix)
{
TrieNode *node = &root;
for (int i = 0; i < prefix.size(); i++)
{
if (node->next[prefix[i] - 'a'] != NULL)
node = node->next[prefix[i] - 'a'];
else
return false;
}
return true;
}
};
```
## 选项
### A
```cpp
struct TrieNode
{
bool isEnd = false;
TrieNode *children[26] = {NULL};
};
class Trie
{
public:
Trie() : root(new TrieNode) {}
void insert(string word)
{
TrieNode *p = root;
for (int i = 0; i < word.size(); ++i)
{
int idx = word[i] - 'a';
if (p->children[idx] == NULL)
p->children[idx] = new TrieNode;
p = p->children[idx];
}
p->isEnd = true;
}
bool search(string word)
{
auto p = root;
for (int i = 0; i < word.size(); ++i)
{
int idx = word[i] - 'a';
if (p->children[idx] == NULL)
return false;
p = p->children[idx];
}
return p->isEnd;
}
bool startsWith(string prefix)
{
auto p = root;
for (int i = 0; i < prefix.size(); ++i)
{
int idx = prefix[i] - 'a';
if (p->children[idx] == NULL)
return false;
p = p->children[idx];
}
return true;
}
private:
TrieNode *root;
};
```
### B
```cpp
struct TrieNode
{
const static int MaxBranchNum = 26;
char data;
bool isEnd = false;
vector<TrieNode *> children = vector<TrieNode *>(26);
TrieNode(char data) : data(data){};
};
class Trie
{
public:
Trie() : root(new TrieNode('/')) {}
void insert(string word)
{
TrieNode *p = root;
for (int i = 0; i < word.size(); ++i)
{
int idx = word[i] - 'a';
if (p->children[idx] == NULL)
p->children[idx] = new TrieNode(word[i]);
p = p->children[idx];
}
p->isEnd = true;
}
bool search(string word)
{
auto p = root;
for (int i = 0; i < word.size(); ++i)
{
int idx = word[i] - 'a';
if (p->children[idx] == NULL)
return false;
p = p->children[idx];
}
return p->isEnd;
}
bool startsWith(string prefix)
{
auto p = root;
for (int i = 0; i < prefix.size(); ++i)
{
int idx = prefix[i] - 'a';
if (p->children[idx] == NULL)
return false;
p = p->children[idx];
}
return true;
}
private:
TrieNode *root;
};
```
### C
```cpp
struct Node
{
map<char, Node *> next;
bool f = false;
};
class Trie
{
public:
Node *head;
/** Initialize your data structure here. */
Trie()
{
head = new Node();
}
/** Inserts a word into the trie. */
void insert(string word)
{
Node *t = head;
for (char c : word)
{
if (t->next[c] == NULL)
{
Node *n = new Node;
t->next[c] = n;
}
t = t->next[c];
}
t->f = true;
}
/** Returns if the word is in the trie. */
bool search(string word)
{
Node *t = head;
for (char c : word)
{
if (t->next[c] == NULL)
{
return false;
}
t = t->next[c];
}
return t->f;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
bool startsWith(string prefix)
{
Node *t = head;
for (char c : prefix)
{
if (t->next[c] == NULL)
{
return false;
}
t = t->next[c];
}
return true;
}
};
```
data/3.算法高阶/5.leetcode-设计/8.535-TinyURL 的加密与解密/config.json 0 → 100644
浏览文件 @ 66b8b16b
{
"node_id": "algorithm-c0fc6a20b2914daba094b822d884d76f",
"keywords": [
"leetcode",
"TinyURL 的加密与解密"
],
"children": [],
"export": [
"solution.json"
],
"title": "TinyURL 的加密与解密"
}
\ No newline at end of file
data/3.算法高阶/5.leetcode-设计/8.535-TinyURL 的加密与解密/desc.html 0 → 100644
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<p>TinyURL是一种URL简化服务， 比如：当你输入一个URL&nbsp;<code>https://leetcode.com/problems/design-tinyurl</code>&nbsp;时，它将返回一个简化的URL&nbsp;<code>http://tinyurl.com/4e9iAk</code>.</p>
<p>要求：设计一个 TinyURL 的加密&nbsp;<code>encode</code>&nbsp;和解密&nbsp;<code>decode</code>&nbsp;的方法。你的加密和解密算法如何设计和运作是没有限制的，你只需要保证一个URL可以被加密成一个TinyURL，并且这个TinyURL可以用解密方法恢复成原本的URL。</p>
data/3.算法高阶/5.leetcode-设计/8.535-TinyURL 的加密与解密/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "4bb1610f42434348b3ea6b82290f04d3"
}
\ No newline at end of file
data/3.算法高阶/5.leetcode-设计/8.535-TinyURL 的加密与解密/solution.md 0 → 100644
浏览文件 @ 66b8b16b
# TinyURL 的加密与解密
<p>TinyURL是一种URL简化服务， 比如：当你输入一个URL&nbsp;<code>https://leetcode.com/problems/design-tinyurl</code>&nbsp;时，它将返回一个简化的URL&nbsp;<code>http://tinyurl.com/4e9iAk</code>.</p>
<p>要求：设计一个 TinyURL 的加密&nbsp;<code>encode</code>&nbsp;和解密&nbsp;<code>decode</code>&nbsp;的方法。你的加密和解密算法如何设计和运作是没有限制的，你只需要保证一个URL可以被加密成一个TinyURL，并且这个TinyURL可以用解密方法恢复成原本的URL。</p>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
// Your Solution object will be instantiated and called as such:
// Solution solution;
// solution.decode(solution.encode(url));
```
## 答案
```cpp
class Solution
{
public:
map<string, string> mp;
string encode(string longUrl)
{
string str = "tinyurl" + to_string(key);
mp[str] = longUrl;
return str;
}
string decode(string shortUrl)
{
return mp[shortUrl];
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
unordered_map<string, string> hashMap;
string TINYURL_PREFIX = "http://tinyurl.com/";
string INDEX = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
string encode(string longUrl)
{
string newShortStr = "";
while (true)
{
for (int i = 0; i < 6; ++i)
{
newShortStr += INDEX[rand() % 62];
}
string resStr = TINYURL_PREFIX + newShortStr;
if (hashMap.count(resStr) == 0)
{
hashMap[resStr] = longUrl;
return resStr;
}
}
}
string decode(string shortUrl)
{
return hashMap[shortUrl];
}
};
```
### B
```cpp
class Solution
{
public:
using ull = unsigned long long;
const ull base = 11;
const string code = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ull len = code.size();
unordered_map<string, string> m;
ull hashCode(string url)
{
ull hash = 0;
for (auto c : url)
{
hash *= base;
hash += c;
}
return hash;
}
string hashToString(ull n)
{
string ans;
while (n)
{
ans += code[n % len];
n /= len;
}
return ans;
}
string encode(string longUrl)
{
string shortUrl = hashToString(hashCode(longUrl));
m[shortUrl] = longUrl;
return shortUrl;
}
string decode(string shortUrl)
{
return m[shortUrl];
}
};
```
### C
```cpp
class Solution
{
public:
unordered_map<string, string> mp;
int i = 0;
string encode(string longUrl)
{
mp[to_string(i)] = longUrl;
string res = "http://tinyurl.com/" + to_string(i++);
return res;
}
string decode(string shortUrl)
{
return mp[shortUrl.substr(19, shortUrl.size() - 19)];
}
};
```
data/3.算法高阶/8.leetcode-贪心/10.330-按要求补齐数组/config.json 0 → 100644
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{
"node_id": "algorithm-877445164cb3460c98afa991e4cba2d9",
"keywords": [
"leetcode",
"按要求补齐数组"
],
"children": [],
"export": [
"solution.json"
],
"title": "按要求补齐数组"
}
\ No newline at end of file
data/3.算法高阶/8.leetcode-贪心/10.330-按要求补齐数组/desc.html 0 → 100644
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<p>给定一个已排序的正整数数组 <em>nums，</em>和一个正整数&nbsp;<em>n 。</em>从&nbsp;<code>[1, n]</code>&nbsp;区间内选取任意个数字补充到&nbsp;<em>nums&nbsp;</em>中，使得&nbsp;<code>[1, n]</code>&nbsp;区间内的任何数字都可以用&nbsp;<em>nums&nbsp;</em>中某几个数字的和来表示。请输出满足上述要求的最少需要补充的数字个数。</p>
<p><strong>示例&nbsp;1:</strong></p>
<pre><strong>输入: </strong><em>nums</em> = <code>[1,3]</code>, <em>n</em> = <code>6</code>
<strong>输出: </strong>1
<strong>解释:</strong>
根据<em> nums&nbsp;</em>里现有的组合&nbsp;<code>[1], [3], [1,3]</code>，可以得出&nbsp;<code>1, 3, 4</code>。
现在如果我们将&nbsp;<code>2</code>&nbsp;添加到&nbsp;<em>nums 中，</em>&nbsp;组合变为: <code>[1], [2], [3], [1,3], [2,3], [1,2,3]</code>。
其和可以表示数字&nbsp;<code>1, 2, 3, 4, 5, 6</code>，能够覆盖&nbsp;<code>[1, 6]</code>&nbsp;区间里所有的数。
所以我们最少需要添加一个数字。</pre>
<p><strong>示例 2:</strong></p>
<pre><strong>输入: </strong><em>nums</em> = <code>[1,5,10]</code>, <em>n</em> = <code>20</code>
<strong>输出:</strong> 2
<strong>解释: </strong>我们需要添加&nbsp;<code>[2, 4]</code>。
</pre>
<p><strong>示例&nbsp;3:</strong></p>
<pre><strong>输入: </strong><em>nums</em> = <code>[1,2,2]</code>, <em>n</em> = <code>5</code>
<strong>输出:</strong> 0
</pre>
data/3.算法高阶/8.leetcode-贪心/10.330-按要求补齐数组/solution.json 0 → 100644
浏览文件 @ 66b8b16b
{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f2e948669b074e0894d025effeeddcb7"
}
\ No newline at end of file
data/3.算法高阶/8.leetcode-贪心/10.330-按要求补齐数组/solution.md 0 → 100644
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# 按要求补齐数组
<p>给定一个已排序的正整数数组 <em>nums，</em>和一个正整数&nbsp;<em>n 。</em>从&nbsp;<code>[1, n]</code>&nbsp;区间内选取任意个数字补充到&nbsp;<em>nums&nbsp;</em>中，使得&nbsp;<code>[1, n]</code>&nbsp;区间内的任何数字都可以用&nbsp;<em>nums&nbsp;</em>中某几个数字的和来表示。请输出满足上述要求的最少需要补充的数字个数。</p>
<p><strong>示例&nbsp;1:</strong></p>
<pre><strong>输入: </strong><em>nums</em> = <code>[1,3]</code>, <em>n</em> = <code>6</code>
<strong>输出: </strong>1
<strong>解释:</strong>
根据<em> nums&nbsp;</em>里现有的组合&nbsp;<code>[1], [3], [1,3]</code>，可以得出&nbsp;<code>1, 3, 4</code>。
现在如果我们将&nbsp;<code>2</code>&nbsp;添加到&nbsp;<em>nums 中，</em>&nbsp;组合变为: <code>[1], [2], [3], [1,3], [2,3], [1,2,3]</code>。
其和可以表示数字&nbsp;<code>1, 2, 3, 4, 5, 6</code>，能够覆盖&nbsp;<code>[1, 6]</code>&nbsp;区间里所有的数。
所以我们最少需要添加一个数字。</pre>
<p><strong>示例 2:</strong></p>
<pre><strong>输入: </strong><em>nums</em> = <code>[1,5,10]</code>, <em>n</em> = <code>20</code>
<strong>输出:</strong> 2
<strong>解释: </strong>我们需要添加&nbsp;<code>[2, 4]</code>。
</pre>
<p><strong>示例&nbsp;3:</strong></p>
<pre><strong>输入: </strong><em>nums</em> = <code>[1,2,2]</code>, <em>n</em> = <code>5</code>
<strong>输出:</strong> 0
</pre>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<int> nums = {1, 5, 10};
int n = 20;
int res = sol.minPatches(nums, n);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
int minPatches(vector<int> &nums, int n)
{
long sum = 0;
int reti = 0;
int i = 0;
while (sum < n)
{
if (i >= nums.size())
break;
if ((sum + 1) >= nums[i])
{
sum += nums[++i];
}
else
{
int temp = sum + 1;
sum += temp;
reti++;
}
}
while (sum < n)
{
int temp = sum + 1;
sum += temp;
}
return reti;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int minPatches(vector<int> &nums, int n)
{
int resultCnt = 0;
int numsSize = nums.size(), index = 0;
long long head = 1;
while (head <= n)
{
if (index < numsSize && nums[index] <= head)
{
head += nums[index++];
}
else
{
head += head;
resultCnt += 1;
}
}
return resultCnt;
}
};
```
### B
```cpp
class Solution
{
public:
int minPatches(vector<int> &nums, int n)
{
int patchCount = 0, i = 0;
long miss = 1;
while (miss <= n)
{
if (i < nums.size() && miss >= nums[i])
miss += nums[i++];
else
{
patchCount++;
miss = miss * 2;
}
}
return patchCount;
}
};
```
### C
```cpp
class Solution
{
public:
int minPatches(vector<int> &nums, int n)
{
int len = nums.size(), i = 0;
long current_coverage = 0;
int ans = 0;
while (current_coverage < n)
{
if (i < len)
{
if (nums[i] <= current_coverage + 1)
{
current_coverage += nums[i];
i++;
}
else
{
ans++;
current_coverage += current_coverage + 1;
}
}
else
{
ans++;
current_coverage += current_coverage + 1;
}
}
return ans;
}
};
```
data/3.算法高阶/8.leetcode-贪心/7.179-最大数/config.json 0 → 100644
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{
"node_id": "algorithm-79927151f5264497aa7cf4054ca7c573",
"keywords": [
"leetcode",
"最大数"
],
"children": [],
"export": [
"solution.json"
],
"title": "最大数"
}
\ No newline at end of file
data/3.算法高阶/8.leetcode-贪心/7.179-最大数/desc.html 0 → 100644
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<p>给定一组非负整数 <code>nums</code>，重新排列每个数的顺序（每个数不可拆分）使之组成一个最大的整数。</p>
<p><strong>注意：</strong>输出结果可能非常大，所以你需要返回一个字符串而不是整数。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入<code>：</code></strong><code>nums = [10,2]</code>
<strong>输出：</strong><code>"210"</code></pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入<code>：</code></strong><code>nums = [3,30,34,5,9]</code>
<strong>输出：</strong><code>"9534330"</code>
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入<code>：</code></strong>nums = [1]
<strong>输出：</strong>"1"
</pre>
<p><strong>示例 4：</strong></p>
<pre>
<strong>输入<code>：</code></strong>nums = [10]
<strong>输出：</strong>"10"
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= nums.length <= 100</code></li>
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
</ul>
data/3.算法高阶/8.leetcode-贪心/7.179-最大数/solution.cpp 0 → 100644
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class Solution
{
public:
bool static cmp(string a, string b)
{
return (a + b) > (b + a); //当a+b>b+a时a排在b前面
}
string largestNumber(vector<int> &nums)
{
int flag = 0;
vector<string> v;
for (vector<int>::iterator i = nums.begin(); i < nums.end(); i++)
{
v.push_back(to_string(*i));
if (*i)
flag = 1;
}
if (!flag)
return "0";
sort(v.begin(), v.end(), cmp); //没有生成实例就要调用，故cmp应声明为static
string s = "";
for (int i = 0; i < v.size(); i++)
s = s + v[i];
return s;
}
};
\ No newline at end of file
data/3.算法高阶/8.leetcode-贪心/7.179-最大数/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "8ffe9f362cee4808887feae6a1a4baab"
}
\ No newline at end of file
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# 最大数
<p>给定一组非负整数 <code>nums</code>，重新排列每个数的顺序（每个数不可拆分）使之组成一个最大的整数。</p>
<p><strong>注意：</strong>输出结果可能非常大，所以你需要返回一个字符串而不是整数。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入<code>：</code></strong><code>nums = [10,2]</code>
<strong>输出：</strong><code>"210"</code></pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入<code>：</code></strong><code>nums = [3,30,34,5,9]</code>
<strong>输出：</strong><code>"9534330"</code>
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入<code>：</code></strong>nums = [1]
<strong>输出：</strong>"1"
</pre>
<p><strong>示例 4：</strong></p>
<pre>
<strong>输入<code>：</code></strong>nums = [10]
<strong>输出：</strong>"10"
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= nums.length <= 100</code></li>
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<int> nums = {3, 30, 34, 5, 9};
string res = sol.largestNumber(nums);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
string ans;
for (int i = 0; i < nums.size(); i++)
{
for (int j = nums.size() - 1; j > i; j--)
{
if (to_string(nums[j - 1]) + to_string(nums[j]) > to_string(nums[j]) + to_string(nums[j - 1]))
{
int tmp = nums[j - 1];
nums[j - 1] = nums[j];
nums[j] = tmp;
}
}
ans += to_string(nums[i]);
}
if (ans + "0" == "0" + ans)
return "0";
return ans;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
string ans;
sort(nums.begin(), nums.end(), [](int a, int b)
{ return to_string(a) + to_string(b) > to_string(b) + to_string(a); });
for (auto it : nums)
{
ans += to_string(it);
}
return ans[0] == '0' ? "0" : ans;
}
};
```
### B
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
if (*max_element(nums.begin(), nums.end()) == 0)
return to_string(0);
vector<string> vec(nums.size());
for (int i = 0; i < nums.size(); ++i)
vec[i] = to_string(nums[i]);
sort(vec.begin(), vec.end(), [](const string &s1, const string &s2)
{ return s1 + s2 > s2 + s1; });
return accumulate(vec.begin(), vec.end(), string());
}
};
```
### C
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
string res = "";
int n = nums.size();
vector<string> tmp;
for (int i = 0; i < n; i++)
{
tmp.push_back(to_string(nums[i]));
}
sort(tmp.begin(), tmp.end(), compare);
for (int i = 0; i < tmp.size(); i++)
{
res += tmp[i];
}
if ('0' == res[0])
{
return "0";
}
else
{
return res;
}
}
};
```
data/3.算法高阶/8.leetcode-贪心/8.316-去除重复字母/config.json 0 → 100644
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{
"node_id": "algorithm-b19a2b60d98244eb9e56813921864871",
"keywords": [
"leetcode",
"去除重复字母"
],
"children": [],
"export": [
"solution.json"
],
"title": "去除重复字母"
}
\ No newline at end of file
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<p>给你一个字符串 <code>s</code> ，请你去除字符串中重复的字母，使得每个字母只出现一次。需保证 <strong>返回结果的字典序最小</strong>（要求不能打乱其他字符的相对位置）。</p>
<p><strong>注意：</strong>该题与 1081 <a href="https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters">https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters</a> 相同</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong><code>s = "bcabc"</code>
<strong>输出<code>：</code></strong><code>"abc"</code>
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong><code>s = "cbacdcbc"</code>
<strong>输出：</strong><code>"acdb"</code></pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= s.length <= 10<sup>4</sup></code></li>
<li><code>s</code> 由小写英文字母组成</li>
</ul>
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "17a3169f78f74c65ae53df020a642995"
}
\ No newline at end of file
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# 去除重复字母
<p>给你一个字符串 <code>s</code> ，请你去除字符串中重复的字母，使得每个字母只出现一次。需保证 <strong>返回结果的字典序最小</strong>（要求不能打乱其他字符的相对位置）。</p>
<p><strong>注意：</strong>该题与 1081 <a href="https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters">https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters</a> 相同</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong><code>s = "bcabc"</code>
<strong>输出<code>：</code></strong><code>"abc"</code>
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong><code>s = "cbacdcbc"</code>
<strong>输出：</strong><code>"acdb"</code></pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= s.length <= 10<sup>4</sup></code></li>
<li><code>s</code> 由小写英文字母组成</li>
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
string s = "cbacdcbc";
string res = sol.removeDuplicateLetters(s);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
string removeDuplicateLetters(string s)
{
int count[26] = {0};
for (int i = 0; i < s.length(); i++)
count[s[i] - 'a']++;
int pos = 0;
for (int i = 0; i < s.length(); i++)
{
if (s[i] < s[pos])
pos = i;
if (count[s[i] - 'a'] == 0)
break;
}
string suffix;
if (pos + 1 > s.length())
suffix = "";
else
suffix = s.substr(pos + 1, s.length() - pos - 1);
return s.length() == 0 ? "" : s[pos] + removeDuplicateLetters(removeOneLetter(suffix, s[pos]));
}
string removeOneLetter(string s, char ch)
{
while (1)
{
int index = s.find(ch);
if (index != -1)
s = s.erase(index, 1);
else
break;
}
return s;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
string removeDuplicateLetters(string s)
{
unordered_map<char, int> mp;
unordered_map<char, bool> isIn;
int len = s.length();
for (int i = 0; i < len; i++)
{
mp[s[i]]++;
isIn[s[i]] = false;
}
stack<char> ans;
for (int i = 0; i < len; i++)
{
if (ans.empty())
{
ans.push(s[i]);
isIn[s[i]] = true;
}
else if (!isIn[s[i]])
{
while (!ans.empty() && s[i] < ans.top())
{
if (mp[ans.top()])
{
isIn[ans.top()] = false;
ans.pop();
}
else
break;
}
ans.push(s[i]);
isIn[s[i]] = true;
}
mp[s[i]]--;
}
string ansS;
while (!ans.empty())
{
ansS += ans.top();
ans.pop();
}
reverse(ansS.begin(), ansS.end());
return ansS;
}
};
```
### B
```cpp
class Solution
{
public:
string removeDuplicateLetters(string s)
{
string ans;
for (int i = 0; i < s.size(); i++)
{
if (ans.find(s[i]) != string::npos)
continue;
for (int j = ans.length() - 1; j >= 0; j--)
{
if (ans[j] > s[i] && s.find(ans[j], i + 1) != string::npos)
{
ans.erase(j);
continue;
}
break;
}
ans += s[i];
}
return ans;
}
};
```
### C
```cpp
class Solution
{
public:
string res = "0";
string removeDuplicateLetters(string s)
{
int count[26] = {0};
int visited[26] = {0};
for (int i = 0; i < s.size(); ++i)
++count[s[i] - 'a'];
for (const auto &c : s)
{
--count[c - 'a'];
if (visited[c - 'a'] == 1)
continue;
while (c < res.back() && count[res.back() - 'a'])
{
visited[res.back() - 'a'] = 0;
res.pop_back();
}
res += c;
visited[c - 'a'] = 1;
}
return res.substr(1);
}
};
```
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{
"node_id": "algorithm-476b83ac859c40f3a6efd6c3632f1ad2",
"keywords": [
"leetcode",
"拼接最大数"
],
"children": [],
"export": [
"solution.json"
],
"title": "拼接最大数"
}
\ No newline at end of file
data/3.算法高阶/8.leetcode-贪心/9.321-拼接最大数/desc.html 0 → 100644
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<p>给定长度分别为&nbsp;<code>m</code>&nbsp;和&nbsp;<code>n</code>&nbsp;的两个数组，其元素由&nbsp;<code>0-9</code>&nbsp;构成，表示两个自然数各位上的数字。现在从这两个数组中选出 <code>k (k &lt;= m + n)</code>&nbsp;个数字拼接成一个新的数，要求从同一个数组中取出的数字保持其在原数组中的相对顺序。</p>
<p>求满足该条件的最大数。结果返回一个表示该最大数的长度为&nbsp;<code>k</code>&nbsp;的数组。</p>
<p><strong>说明: </strong>请尽可能地优化你算法的时间和空间复杂度。</p>
<p><strong>示例&nbsp;1:</strong></p>
<pre><strong>输入:</strong>
nums1 = <code>[3, 4, 6, 5]</code>
nums2 = <code>[9, 1, 2, 5, 8, 3]</code>
k = <code>5</code>
<strong>输出:</strong>
<code>[9, 8, 6, 5, 3]</code></pre>
<p><strong>示例 2:</strong></p>
<pre><strong>输入:</strong>
nums1 = <code>[6, 7]</code>
nums2 = <code>[6, 0, 4]</code>
k = <code>5</code>
<strong>输出:</strong>
<code>[6, 7, 6, 0, 4]</code></pre>
<p><strong>示例 3:</strong></p>
<pre><strong>输入:</strong>
nums1 = <code>[3, 9]</code>
nums2 = <code>[8, 9]</code>
k = <code>3</code>
<strong>输出:</strong>
<code>[9, 8, 9]</code></pre>
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d4bef81d1bd84c0db36d6488b1932479"
}
\ No newline at end of file
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# 拼接最大数
<p>给定长度分别为&nbsp;<code>m</code>&nbsp;和&nbsp;<code>n</code>&nbsp;的两个数组，其元素由&nbsp;<code>0-9</code>&nbsp;构成，表示两个自然数各位上的数字。现在从这两个数组中选出 <code>k (k &lt;= m + n)</code>&nbsp;个数字拼接成一个新的数，要求从同一个数组中取出的数字保持其在原数组中的相对顺序。</p>
<p>求满足该条件的最大数。结果返回一个表示该最大数的长度为&nbsp;<code>k</code>&nbsp;的数组。</p>
<p><strong>说明: </strong>请尽可能地优化你算法的时间和空间复杂度。</p>
<p><strong>示例&nbsp;1:</strong></p>
<pre><strong>输入:</strong>
nums1 = <code>[3, 4, 6, 5]</code>
nums2 = <code>[9, 1, 2, 5, 8, 3]</code>
k = <code>5</code>
<strong>输出:</strong>
<code>[9, 8, 6, 5, 3]</code></pre>
<p><strong>示例 2:</strong></p>
<pre><strong>输入:</strong>
nums1 = <code>[6, 7]</code>
nums2 = <code>[6, 0, 4]</code>
k = <code>5</code>
<strong>输出:</strong>
<code>[6, 7, 6, 0, 4]</code></pre>
<p><strong>示例 3:</strong></p>
<pre><strong>输入:</strong>
nums1 = <code>[3, 9]</code>
nums2 = <code>[8, 9]</code>
k = <code>3</code>
<strong>输出:</strong>
<code>[9, 8, 9]</code></pre>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<int> nums1 = {3, 4, 6, 5};
vector<int> nums2 = {9, 1, 2, 5, 8, 3};
int k = 5;
vector<int> res = sol.maxNumber(nums1, nums2, k);
for (auto i : res)
cout << i << " ";
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
struct node
{
int val;
int vec;
int pos;
};
static bool cmp(node a, node b)
{
return a.val > b.val;
}
vector<int> maxNumber(vector<int> &nums1, vector<int> &nums2, int k)
{
int n = nums1.size();
int m = nums2.size();
struct node all[m + n];
vector<int> ans;
for (int i = 0; i < n; i++)
{
all[i].vec = 1;
all[i].val = nums1[i];
all[i].pos = i;
}
for (int i = 0; i < m; i++)
{
all[i + n].vec = 2;
all[i + n].val = nums2[i];
all[i + n].pos = i;
}
sort(all, all + m + n, cmp);
int now1 = 0;
int now2 = 0;
int st = 0;
while (st != n + 1)
{
for (int i = 0; i < n + m; i++)
{
if (all[i].pos >= now1 && n - all[i].pos + m - now2 >= k - st)
{
ans.push_back(all[i].val);
now1 = all[i].pos + 1;
st++;
break;
}
}
}
return ans;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
vector<int> getNumsFromVector(vector<int> &nums, int k)
{
int n = nums.size();
if (n == 0)
return {};
vector<int> stack;
stack.push_back(nums[0]);
for (int i = 1; i < n; ++i)
{
while (!stack.empty() && nums[i] > stack.back() && (n - i > k - (int)stack.size()))
{
stack.pop_back();
}
stack.push_back(nums[i]);
}
if (stack.size() > k)
{
vector<int> res(stack.begin(), stack.begin() + k);
return res;
}
return stack;
}
int maxNumsVector(vector<int> &nums1, int index1, vector<int> &nums2, int index2)
{
int m = nums1.size(), n = nums2.size();
if (m == 0)
return 2;
if (n == 0)
return 1;
int p1 = index1, p2 = index2;
for (; p1 < m && p2 < n; ++p1, ++p2)
{
if (nums1[p1] < nums2[p2])
{
return 2;
}
if (nums1[p1] > nums2[p2])
{
return 1;
}
}
if (p1 < m)
{
return 1;
}
else
{
return 2;
}
}
vector<int> merge(vector<int> &nums1, vector<int> &nums2)
{
int m = nums1.size(), n = nums2.size();
if (m == 0)
return nums2;
if (n == 0)
return nums1;
vector<int> res(m + n, 0);
int p1 = 0, p2 = 0;
int i = 0;
while (p1 < m && p2 < n)
{
if (nums1[p1] < nums2[p2])
{
res[i++] = nums2[p2++];
}
else if (nums1[p1] == nums2[p2])
{
if (maxNumsVector(nums1, p1, nums2, p2) == 1)
{
res[i++] = nums1[p1++];
}
else
{
res[i++] = nums2[p2++];
}
}
else
{
res[i++] = nums1[p1++];
}
}
while (p1 < m)
{
res[i++] = nums1[p1++];
}
while (p2 < n)
{
res[i++] = nums2[p2++];
}
return res;
}
vector<int> maxNumber(vector<int> &nums1, vector<int> &nums2, int k)
{
int m = nums1.size();
int n = nums2.size();
int begin = max(0, k - n), end = min(k, m);
vector<int> ansNums(k, 0);
for (int i = begin; i <= end; ++i)
{
vector<int> tempNums1 = getNumsFromVector(nums1, i);
vector<int> tempNums2 = getNumsFromVector(nums2, k - i);
vector<int> mergeRes = merge(tempNums1, tempNums2);
if (maxNumsVector(ansNums, 0, mergeRes, 0) == 2)
{
ansNums = mergeRes;
}
}
return ansNums;
}
};
```
### B
```cpp
class Solution
{
public:
vector<int> find_kMax(vector<int> &num, int k)
{
vector<int> res;
int i = 0;
while (k > 0)
{
int idx = i;
for (; i < num.size() - k + 1; ++i)
{
if (num[i] > num[idx])
idx = i;
}
res.push_back(num[idx]);
i = idx + 1;
--k;
}
return res;
}
vector<int> merge(vector<int> &nums1, vector<int> &nums2)
{
vector<int> res;
for (auto iter1 = nums1.begin(), iter2 = nums2.begin(); iter1 != nums1.end() || iter2 != nums2.end();)
{
if (lexicographical_compare(iter1, nums1.end(), iter2, nums2.end()))
res.push_back(*iter2++);
else
res.push_back(*iter1++);
}
return res;
}
vector<int> maxNumber(vector<int> &nums1, vector<int> &nums2, int k)
{
vector<int> res;
for (int i = 0; i <= k; ++i)
{
int j = k - i;
if (i > nums1.size() || j > nums2.size())
continue;
auto tmp1 = find_kMax(nums1, i);
auto tmp2 = find_kMax(nums2, j);
res = max(res, merge(tmp1, tmp2));
}
return res;
}
};
```
### C
```cpp
class Solution
{
public:
vector<int> maxNumber(vector<int> &nums1, vector<int> &nums2, int k)
{
int m = nums1.size(), n = nums2.size();
vector<int> res;
for (int i = max(0, k - n); i <= min(k, m); i++)
res = max(res, mergeVector(maxVector(nums1, i), maxVector(nums2, k - i)));
return res;
}
vector<int> maxVector(vector<int> nums, int k)
{
int drop = nums.size() - k;
vector<int> res;
for (int num : nums)
{
while (drop && res.size() && res.back() < num)
{
res.pop_back();
--drop;
}
res.push_back(num);
}
res.resize(k);
return res;
}
vector<int> mergeVector(vector<int> nums1, vector<int> nums2)
{
vector<int> res;
while (nums1.size() + nums2.size())
{
vector<int> &tmp = nums1 > nums2 ? nums1 : nums2;
res.push_back(tmp[0]);
tmp.erase(tmp.begin());
}
return res;
}
};
```
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "c5fd58daecf5409e85df121fe9e7110b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "30dc3fe246f543ec9f26f44b0051cab1"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "7026e2bc9ba945ecb75693f21ee96a59"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "c768955a81964a16af533b1180acea75"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "9fd7021efe434546947e98c9df345ff4"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2d7778e134004e3488e6750325998ea7"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "c2ed088169324224808baa01bbf3ead3"
}
\ No newline at end of file
data_backup/1.leetcode/1007-行相等的最少多米诺旋转/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "609341761cd74b8485fe88a08430b237"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "05818cc221d4482e8312716a8e64ab0e"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "02026ccbe5a949de8e9919d285864050"
}
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data_backup/1.leetcode/101-对称二叉树/solution.md
浏览文件 @ 66b8b16b
# 对称二叉树
<p>给定一个二叉树，检查它是否是镜像对称的。</p>
<p>&nbsp;</p>
......@@ -30,10 +31,22 @@
<p>你可以运用递归和迭代两种方法解决这个问题吗？</p>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
......@@ -42,21 +55,110 @@
## 答案
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
if (!root)
return true;
queue<TreeNode *> q;
q.push(root->left);
q.push(root->right);
while (!q.empty())
{
auto p1 = q.front();
q.pop();
auto p2 = q.front();
q.pop();
if (!p1 && !p2)
continue;
if (!p1 || !p2)
return false;
if (p1->val != p2->val)
return false;
q.push(p1->right);
q.push(p1->left);
q.push(p2->right);
q.push(p2->left);
}
return true;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
queue<TreeNode *> q1, q2;
if (!root)
return true;
q1.push(root->left);
q2.push(root->right);
while (!q1.empty() && !q2.empty())
{
TreeNode *node1 = q1.front();
q1.pop();
TreeNode *node2 = q2.front();
q2.pop();
if (!node1 && !node2)
continue;
if ((!node1 && node2) || (node1 && !node2) || (node1->val != node2->val))
return false;
q1.push(node1->left);
q1.push(node1->right);
q2.push(node2->right);
q2.push(node2->left);
}
return true;
}
};
```
### B
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
if (!root)
return true;
return helper(root->left, root->right);
}
bool helper(TreeNode *left, TreeNode *right)
{
if (!left && !right)
return true;
if ((!left && right) || (left && !right) || (left->val != right->val))
return false;
return helper(left->left, right->right) && helper(left->right, right->left);
}
};
```
### C
```cpp
class Solution
{
public:
bool isSymmetric(TreeNode *root)
{
return ismirror(root, root);
}
bool ismirror(TreeNode *t1, TreeNode *t2)
{
if (t1 == NULL && t2 == NULL)
return true;
if (t1 == NULL || t2 == NULL)
return false;
return (t1->val == t2->val) && ismirror(t1->left, t2->right) && ismirror(t1->right, t2->left);
}
};
```
data_backup/1.leetcode/1010-总持续时间可被 60 整除的歌曲/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "aae782a36ce34852a0c7e3c73191a549"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "925fa09a52a44aabbbccf6e83911bf46"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fe1341408a504f5daae4ebf14ae299f6"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "9bd9496e3de44835b70e152f2120d974"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6f8984c5ec194d4bb961c5e9e12c5ffd"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "30b0b6f6226d4abcaa84b6c406768769"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fb1e6aeafb0744928925929af214b4db"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0aefe9665b4243c086f30c8f1b0c1d35"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "21c3b557b87e46899b0e52caaf2b6894"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "7aaf537ab5b44771a2c4bbc8d2cf6526"
}
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data_backup/1.leetcode/102-二叉树的层序遍历/solution.md
浏览文件 @ 66b8b16b
# 二叉树的层序遍历
<p>给你一个二叉树，请你返回其按 <strong>层序遍历</strong> 得到的节点值。 （即逐层地，从左到右访问所有节点）。</p>
<p> </p>
......@@ -25,10 +26,22 @@
</pre>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
......@@ -37,21 +50,147 @@
## 答案
```cpp
class Solution
{
public:
vector<vector<int>> levelOrder(TreeNode *root)
{
vector<vector<int>> res;
queue<TreeNode *> q;
if (root != NULL)
{
while (!q.empty())
{
int size = q.size();
vector<int> temp;
for (int i = 0; i < size; i++)
{
TreeNode *t = q.front();
q.pop();
temp.push_back(t->val);
if (t->left != NULL)
q.push(t->left);
if (t->right != NULL)
q.push(t->right);
}
res.push_back(temp);
temp.clear();
}
}
return res;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
vector<vector<int>> levelOrder(TreeNode *root)
{
vector<vector<int>> ret;
if (!root)
return ret;
queue<TreeNode *> q;
q.push(root);
while (!q.empty())
{
int currNodeSize = q.size();
ret.push_back(vector<int>());
for (int i = 1; i <= currNodeSize; i++)
{
TreeNode *node = q.front();
q.pop();
ret.back().push_back(node->val);
if (node->left)
q.push(node->left);
if (node->right)
q.push(node->right);
}
}
return ret;
}
};
```
### B
```cpp
class Solution
{
public:
int depth(TreeNode *root)
{
if (root == NULL)
return 0;
return max(depth(root->left), depth(root->right)) + 1;
}
void levelOrder(vector<vector<int>> &ans, TreeNode *node, int level)
{
if (!node)
return;
ans[level].push_back(node->val);
levelOrder(ans, node->left, level - 1);
levelOrder(ans, node->right, level - 1);
}
vector<vector<int>> levelOrderBottom(TreeNode *root)
{
int d = depth(root);
vector<vector<int>> ans(d, vector<int>{});
levelOrder(ans, root, d - 1);
return ans;
}
};
```
### C
```cpp
class Solution
{
public:
void dfs(TreeNode *root, vector<vector<int>> &res)
{
queue<TreeNode *> q;
q.push(root);
while (!q.empty())
{
int sz = q.size();
vector<int> tp;
while (sz > 0)
{
TreeNode *p = q.front();
q.pop();
if (p->left != NULL)
{
q.push(p->left);
}
if (p->right != NULL)
{
q.push(p->right);
}
tp.push_back(p->val);
sz--;
}
res.push_back(tp);
}
}
vector<vector<int>> levelOrder(TreeNode *root)
{
vector<vector<int>> res;
if (root == NULL)
{
return res;
}
dfs(root, res);
return res;
}
};
```
data_backup/1.leetcode/1020-飞地的数量/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "783b99867b01438786d8e908915e3868"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6582648f7f4b428eabc19c160a6b105c"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "c941327100fd4953ad6b7d7c0632058b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "ab58c78a3fb94f319eed0c643df90b47"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "c30595a58fbb49088e3fb12fefa5cc29"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "275efc18bde0410293debbb6ca73ea7b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6fd97bfd6e3f49659559ebe411a5b73d"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "74552887a9ea4d239dc0fc1e43eddf30"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "e5a7631e14234838a6e00c656242a134"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "eb464e5a900d40b6a3ec8e6b24511665"
}
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data_backup/1.leetcode/103-二叉树的锯齿形层序遍历/solution.md
浏览文件 @ 66b8b16b
# 二叉树的锯齿形层序遍历
<p>给定一个二叉树，返回其节点值的锯齿形层序遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。</p>
<p>例如：<br />
......@@ -23,10 +24,22 @@
</pre>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
......@@ -35,21 +48,232 @@
## 答案
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
vector<vector<int>> result;
if (root == NULL)
{
return result;
}
deque<TreeNode *> node_queue;
node_queue.push_back(root);
node_queue.push_back(NULL);
deque<int> level_list;
bool is_order_left = true;
while (node_queue.size() > 0)
{
TreeNode *curr_node = node_queue.front();
node_queue.pop_front();
if (curr_node != NULL)
{
if (curr_node->left != NULL)
{
node_queue.push_back(curr_node->left);
}
if (curr_node->right != NULL)
{
node_queue.push_back(curr_node->right);
}
}
else
{
vector<int> tmp;
for (int a : level_list)
{
tmp.push_back(a);
}
result.push_back(tmp);
level_list.clear();
if (node_queue.size() > 0)
{
node_queue.push_back(NULL);
}
is_order_left = !is_order_left;
}
}
return result;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
if (root == NULL)
return {};
vector<vector<int>> res;
int flag = 1;
stack<TreeNode *> a, b;
a.push(root);
while (!a.empty() || !b.empty())
{
vector<int> cur;
while (flag == 1 && !a.empty())
{
root = a.top();
cur.push_back(root->val);
a.pop();
if (root->left != NULL)
b.push(root->left);
if (root->right != NULL)
b.push(root->right);
}
while (flag == -1 && !b.empty())
{
root = b.top();
cur.push_back(root->val);
b.pop();
if (root->right != NULL)
a.push(root->right);
if (root->left != NULL)
a.push(root->left);
}
flag = -1 * flag;
res.push_back(cur);
}
return res;
}
};
```
### B
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
vector<vector<int>> ans;
if (!root)
return ans;
queue<TreeNode *> qnode;
bool orderByLeft = true;
qnode.push(root);
while (!qnode.empty())
{
int levelSize = qnode.size();
deque<int> level;
for (int i = 0; i < levelSize; i++)
{
auto curNode = qnode.front();
qnode.pop();
if (orderByLeft)
{
level.push_back(curNode->val);
}
else
{
level.push_front(curNode->val);
}
if (curNode->left != NULL)
qnode.push(curNode->left);
if (curNode->right != NULL)
qnode.push(curNode->right);
}
orderByLeft = !orderByLeft;
vector<int> curlevel{level.begin(), level.end()};
ans.push_back(curlevel);
}
return ans;
}
};
```
### C
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
if (!root)
return {};
vector<vector<int>> res;
queue<TreeNode *> q{{root}};
int cnt = 0;
while (!q.empty())
{
vector<int> oneLevel;
for (int i = q.size(); i > 0; i--)
{
TreeNode *t = q.front();
q.pop();
oneLevel.push_back(t->val);
if (t->left)
q.push(t->left);
if (t->right)
q.push(t->right);
}
if (cnt % 2 == 1)
reverse(oneLevel.begin(), oneLevel.end());
res.push_back(oneLevel);
cnt++;
}
return res;
}
};
```
### D
```cpp
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode *root)
{
vector<vector<int>> results;
if (root == NULL)
{
return results;
}
zigzagLevelOrderDFS_helper(root, 0, results);
return results;
}
void zigzagLevelOrderDFS_helper(TreeNode *node, int level, vector<vector<int>> &results)
{
if (level >= results.size())
{
vector<int> new_level;
new_level.push_back(node->val);
results.push_back(new_level);
}
else
{
if (level % 2 == 0)
{
results[level].push_back(node->val);
}
else
{
results[level].insert(results[level].begin(), node->val);
}
}
if (node->left != NULL)
{
zigzagLevelOrderDFS_helper(node->left, level + 1, results);
}
if (node->right != NULL)
{
zigzagLevelOrderDFS_helper(node->right, level + 1, results);
}
}
};
```
data_backup/1.leetcode/1030-距离顺序排列矩阵单元格/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d1fa694ebe224546bcf4ce6d013eda9b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "e13470b44c6347608e24ab44ad4d87aa"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fe5793ab8cb8475a90ed10f540ec1a52"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2b36f3ad745d486c9e60e9bc2e42e8f9"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "b184f3ab93b64a12ad574dbe60698365"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "bd8ea3e19f1747a89664be42038106ab"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "35578c946cd54360a33e07eef28895f7"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "ec34b01121b2463ebb648f647f6d062e"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "cf6fe090237247ce95e7dede7106467f"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6ae7cf591ee94960abb05e1c1f66d219"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5f5c2ffafdbf4406a9d6f6b624247445"
}
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"source": "solution.md",
"exercise_id": "24707b7591924930874ca286a3be8e37"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6bc7dee34c624a93a7a9b1a9cb64fecc"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "35343769668c4e31bfe11c649f8f30ed"
}
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{
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"source": "solution.md",
"exercise_id": "df02e26c0dda4345b586826bcb745127"
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "515692181eb54cc3b93ce8c89cc48346"
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{
"type": "code_options",
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"source": "solution.md",
"exercise_id": "d8c6e05a7254467fa4b2870844ba108c"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6eb2b55b68f0458a9cda5ce69bc80284"
}
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "da090a618af849e9a17b84af65408010"
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{
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"source": "solution.md",
"exercise_id": "5c305ff26a86425eb597a170f03d8ce6"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "93a0b9eb8fd048bfbde7fb167751f302"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "8cb6fdb90be44a2d871fb2b2d727c69e"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "4ef77ebc90494b749a3f20ee1ac2fa40"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5f48d10a400144e4909f80c21d65a215"
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......@@ -37,10 +37,22 @@
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
......@@ -49,21 +61,97 @@
## 答案
```cpp
其他三个都是错的
```
## 选项
### A
```cpp
class Solution
{
public:
bool hasPathSum(TreeNode *root, int sum)
{
if (root == NULL)
{
return false;
}
vector<TreeNode *> node_stack;
vector<int> sum_stack;
node_stack.push_back(root);
sum_stack.push_back(sum - root->val);
TreeNode *node;
int curr_sum;
while (!node_stack.empty())
{
node = node_stack.back();
node_stack.pop_back();
curr_sum = sum_stack.back();
sum_stack.pop_back();
if ((node->right == NULL) && (node->left == NULL) && (curr_sum == 0))
{
return true;
}
if (node->right != NULL)
{
node_stack.push_back(node->right);
sum_stack.push_back(curr_sum - node->right->val);
}
if (node->left != NULL)
{
node_stack.push_back(node->left);
sum_stack.push_back(curr_sum - node->left->val);
}
}
return false;
}
};
```
### B
```cpp
class Solution
{
public:
bool hasPathSum(TreeNode *root, int sum)
{
if (root == NULL)
return false;
stack<pair<TreeNode *, int>> s;
s.push(make_pair(root, sum));
TreeNode *node = root;
int currSum;
while (!s.empty())
{
node = s.top().first;
currSum = s.top().second;
s.pop();
if (node->left == NULL && node->right == NULL && node->val == currSum)
return true;
if (node->right)
s.push(make_pair(node->right, currSum - node->val));
if (node->left)
s.push(make_pair(node->left, currSum - node->val));
}
return false;
}
};
```
### C
```cpp
class Solution
{
public:
bool hasPathSum(TreeNode *root, int sum)
{
if (!root)
return false;
if (!root->left && !root->right && root->val == sum)
return true;
return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}
};
```
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "9686849f06194dad92dbab60aa2cee1a"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "243e74999353408a9ec3e8b503d20732"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "93ec174c3021401db653cd25a9240d4d"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d3821c09bf454cfb8cc687e5540cf1e7"
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"type": "code_options",
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"source": "solution.md",
"exercise_id": "9f06d9767c0345ddac93f7031508a286"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0f37190cde4c4c3fad642d36c04c1e5c"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "634cb9e4787a4f1ba9b4c208ce0dd3d7"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d2849892ea524bd9b1df3a99548298ce"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "eb5f0c973a0e4e9d98d16044625d1527"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6eed9e45a2bf4bcf9b2fbb6edb66693a"
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......@@ -41,9 +41,22 @@
</div>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
......@@ -53,21 +66,131 @@
## 答案
```cpp
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode *root, int sum)
{
int pos = 0;
vector<int> path;
vector<vector<int>> result;
dfs(root, sum, pos, path, result);
return result;
}
private:
void dfs(TreeNode *root, int sum, int &pos, vector<int> &path, vector<vector<int>> &result)
{
if (root == NULL)
return;
pos += root->val;
path.push_back(root->val);
if (!root->left && !root->right && pos == sum)
result.push_back(path);
dfs(root->left, sum, pos, path, result);
dfs(root->right, sum, pos, path, result);
path.pop_back();
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
void find(TreeNode *node, vector<vector<int>> &result, vector<int> &path, int sum)
{
path.push_back(node->val);
if (node->left == NULL && node->right == NULL && sum == node->val)
{
result.push_back(path);
return;
}
if (node->left != NULL)
{
find(node->left, result, path, sum - node->val);
path.pop_back();
}
if (node->right != NULL)
{
find(node->right, result, path, sum - node->val);
path.pop_back();
}
}
vector<vector<int>> pathSum(TreeNode *root, int sum)
{
vector<vector<int>> result;
vector<int> path;
if (root == NULL)
{
return result;
}
find(root, result, path, sum);
return result;
}
};
```
### B
```cpp
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode *root, int sum)
{
vector<vector<int>> res;
if (!root)
return res;
vector<int> temp;
dfs(root, res, temp, sum);
return res;
}
void dfs(TreeNode *root, vector<vector<int>> &res, vector<int> temp, int sum)
{
temp.push_back(root->val);
if (!root->left && !root->right && 0 == sum - root->val)
res.push_back(temp);
if (root->left)
dfs(root->left, res, temp, sum - root->val);
if (root->right)
dfs(root->right, res, temp, sum - root->val);
}
};
```
### C
```cpp
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode *root, int targetSum)
{
DFS(root, targetSum);
return result;
}
void DFS(TreeNode *root, int targetSum)
{
if (root == nullptr)
return;
tmp_path.emplace_back(root->val);
targetSum -= root->val;
if (root->left == nullptr && root->right == nullptr && targetSum == 0)
{
result.emplace_back(tmp_path);
}
DFS(root->left, targetSum);
DFS(root->right, targetSum);
tmp_path.pop_back();
}
private:
vector<vector<int>> result;
vector<int> tmp_path;
};
```
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f011dbf399af44298d68db9908a6e7d8"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5359519347ea449fa95ab62f9cef268c"
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "3b96d201f3704f709d931e813774aba4"
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"source": "solution.md",
"exercise_id": "19dee2eaae5244c88559118f5f4b1248"
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5c58a0e700ee4b57abde4c2ba8af47e9"
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5c54f087a9a04dadb2b6de83c91bc4c8"
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "faee8dc60ea441a290be2a74095af09e"
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"exercise_id": "8566298757cf435c813e23daa53101d6"
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"source": "solution.md",
"exercise_id": "8433469531594e32a9f8a7e32616760e"
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "95d6e30a39bf41c3a1bb69275bd177be"
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"source": "solution.md",
"exercise_id": "30213f5e217344ee9d2b46c15e141908"
}
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "33849e7a24eb4dcfbce167da768afdff"
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2c41ec80637a43dc8b5891357445ff79"
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "48e38d64746d49da837e856b32d647d7"
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"exercise_id": "0cfbf57d17fa41f381bb853b60a326f0"
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"type": "code_options",
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}
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"source": "solution.md",
"exercise_id": "b79e5d71b7324c21b0f3f17ba95f9cc9"
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"exercise_id": "0a4b660abfef4583840a1eafd3af7b0a"
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"exercise_id": "b195f55bfc954fe0a97d8da85c1ec5ec"
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"exercise_id": "f97c625ee7974d98bde4a20be287a277"
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"source": "solution.md",
"exercise_id": "8e6e472de08e451fbc6de43bad0e3f1d"
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"exercise_id": "03e879a865b341aea25b0f99fa335f5f"
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"source": "solution.md",
"exercise_id": "70c2a3d90789410dbf28ce5667f442a3"
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"source": "solution.md",
"exercise_id": "14880c5fe7db402892a7c63e295138e8"
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"source": "solution.md",
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"source": "solution.md",
"exercise_id": "c150b2b1151d4315acdceca74ed26f71"
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"source": "solution.md",
"exercise_id": "858d77dcba0c4da09df11287d55cbfe6"
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"author": "CSDN.net",
"source": "solution.md",
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"author": "CSDN.net",
"source": "solution.md",
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"source": "solution.md",
"exercise_id": "a7833dded1654c51a28bbb6404f61e34"
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"source": "solution.md",
"exercise_id": "c99bf3623aa94100a46db697dfb8b8ea"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2ce736ca84c048c09e701125bbc5e47c"
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2bf6d7947e7e44539822918daa2809e5"
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "be9e60f538c8407db676394baa18c53b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "be5e43d004d14e78a4947f61d4dbd569"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d79fe0a78438468eb232b582e11a574e"
}
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"source": "solution.md",
"exercise_id": "b637b9ee199840e498408adb5dca3ece"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "8cdf201d51344481a453469ac082b2c7"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5a879ff8bcfc44abb63678d6ce7fb9cd"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5059fcf396ac415fbabed95ab11058c2"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6a1d92c07a854ed5babfd1b0cbf01223"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "59bebd84991c48a2a3fcf14b0158baa0"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "ff76265c13ae413bb91ac8d9f5486150"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2eb35eaca98b4969bd9edcecb5533a01"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "ae40bdd8717e4d06826b35f2e8eb43ba"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "231d4c4d416e4852b4e4b63f49656ab2"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0465a903c9704e329f0d33550a7df1e1"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "812158f6189d41738ff38e2b523de2e4"
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{
"type": "code_options",
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"source": "solution.md",
"exercise_id": "522c32f51af9407bb7d3e8a0610be338"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f538d906227543018d37757859b46737"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "cd6a5cdca92a4a54b2142e369a516fd9"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0688cbe5ebee4d899f1c2da1b17b49c9"
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data_backup/1.leetcode/126-单词接龙 II/solution.md
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# 单词接龙 II
<p>按字典 <code>wordList</code> 完成从单词 <code>beginWord</code> 到单词 <code>endWord</code> 转化，一个表示此过程的 <strong>转换序列</strong> 是形式上像 <code>beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub></code> 这样的单词序列，并满足：</p>
<div class="original__bRMd">
......@@ -48,33 +49,307 @@
</div>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<vector<string>> res;
string beginWord = "hit", endWord = "cog";
vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};
res = sol.findLadders(beginWord, endWord, wordList);
for (auto i : res)
{
for (auto j : i)
cout << j << " ";
cout << endl;
}
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
unordered_map<string, vector<string>> tree;
vector<vector<string>> ans;
void dfs(vector<string> &cur, string curWord, string endWord)
{
if (curWord == endWord)
{
ans.push_back(cur);
return;
}
for (string s : tree[curWord])
{
cur.push_back(s);
dfs(cur, s, endWord);
cur.pop_back();
}
}
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
if (wordList.size() == 0 || find(wordList.begin(), wordList.end(), endWord) == wordList.end())
return {};
unordered_set<string> bfsFromBegin{beginWord};
unordered_set<string> bfsFromEnd{endWord};
unordered_set<string> dirc(wordList.begin(), wordList.end());
bool findFlag = false, reverseFlag = false;
while (!bfsFromBegin.empty())
{
unordered_set<string> next;
for (string s : bfsFromBegin)
dirc.erase(s);
for (string s1 : bfsFromBegin)
{
for (int i = 0; i < s1.size(); ++i)
{
string s2 = s1;
for (char c = 'a'; c <= 'z'; ++c)
{
s2[i] = c;
if (dirc.count(s2) == 0)
continue;
next.insert(s2);
}
reverseFlag ? tree[s2].push_back(s1) : tree[s1].push_back(s2);
}
}
bfsFromBegin = next;
if (bfsFromBegin.size() > bfsFromEnd.size())
{
reverseFlag = !reverseFlag;
swap(bfsFromBegin, bfsFromEnd);
}
if (findFlag)
break;
}
vector<string> cur = {beginWord};
dfs(cur, beginWord, endWord);
return ans;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
bool differ_one(string &s, string &t)
{
int n = 0;
for (int i = 0; i < s.size(); ++i)
if ((n += s[i] != t[i]) > 1)
return false;
return n == 1;
}
vector<vector<int>> fa;
vector<string> tmp;
vector<vector<string>> ans;
void dfs(int index, vector<string> &wordList)
{
if (fa[index].empty())
{
reverse(tmp.begin(), tmp.end());
ans.push_back(tmp);
reverse(tmp.begin(), tmp.end());
}
for (auto &x : fa[index])
{
tmp.push_back(wordList[x]);
dfs(x, wordList);
tmp.pop_back();
}
}
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
int b = -1, e = -1, x;
for (int i = 0; i < wordList.size(); ++i)
{
if (wordList[i] == beginWord)
b = i;
if (wordList[i] == endWord)
e = i;
}
if (e == -1)
return ans;
if (b == -1)
wordList.push_back(beginWord), b = wordList.size() - 1;
queue<int> que;
fa.assign(wordList.size(), vector<int>());
vector<int> index(wordList.size(), 0);
que.push(b), index[b] = 1;
while (!que.empty())
{
x = que.front(), que.pop();
if (index[e] && index[x] >= index[e])
break;
for (int i = 0; i < wordList.size(); ++i)
if ((index[i] == 0 || index[x] + 1 == index[i]) && differ_one(wordList[x], wordList[i]))
if (index[i] == 0)
index[i] = index[x] + 1, que.push(i), fa[i].push_back(x);
else
fa[i].push_back(x);
}
if (index[e] == 0)
return ans;
tmp.push_back(endWord);
dfs(e, wordList);
tmp.pop_back();
return ans;
}
};
```
### B
```cpp
class Solution
{
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> list(wordList.begin(), wordList.end());
if (!list.count(endWord))
return {};
unordered_map<string, unordered_set<string>> trace;
unordered_set<string> p;
unordered_set<string> q{beginWord};
while (q.size() && !trace.count(endWord))
{
for (auto w : q)
{
list.erase(w);
}
p.clear();
for (auto str : q)
{
for (int i = 0; i < str.size(); ++i)
{
string tmp = str;
for (char c = 'a'; c <= 'z'; ++c)
{
tmp[i] = c;
if (list.find(tmp) == list.end())
continue;
trace[tmp].insert(str);
p.insert(tmp);
}
}
}
q = p;
}
if (!trace.size())
return {};
vector<vector<string>> res;
dfs(res, {}, trace, endWord);
return res;
}
void dfs(vector<vector<string>> &res, vector<string> path, unordered_map<string, unordered_set<string>> &trace, string &str)
{
path.push_back(str);
if (trace.count(str) == 0)
{
reverse(path.begin(), path.end());
res.push_back(path);
return;
}
for (auto word : trace[str])
dfs(res, path, trace, word);
}
};
```
### C
```cpp
class Solution
{
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
{
vector<vector<string>> ans;
unordered_set<string> dict(wordList.begin(), wordList.end());
vector<string> p{beginWord};
queue<vector<string>> paths;
paths.push(p);
unordered_set<string> words;
int level = 1, minlength = INT_MAX;
while (!paths.empty())
{
vector<string> t = paths.front();
paths.pop();
if (t.size() > level)
{
for (string s : words)
dict.erase(s);
words.clear();
level = t.size();
if (level > minlength)
break;
}
string last = t.back();
for (int i = 0; i < last.size(); ++i)
{
string newlast = last;
for (char ch = 'a'; ch <= 'z'; ++ch)
{
newlast[i] = ch;
if (!dict.count(newlast))
continue;
words.insert(newlast);
vector<string> nextpath = t;
nextpath.push_back(newlast);
if (newlast == endWord)
{
ans.push_back(nextpath);
minlength = level;
}
else
{
paths.push(nextpath);
}
}
}
}
return ans;
}
};
```
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "29ef3c5a668b457cac60f2c6fb4844de"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "532a0b44a27a422c856ba674fde7ef44"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "1b7860a8b6044744b8908086824d9831"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "8b62124912b840d9975a8511dc24d6f2"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "91702b08a4eb41de9cb30d0e9d51c696"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "110eaed10a434c6c9193ffbcfd699c33"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "eb94f65ace2c4dc2bb685c77189d2de2"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6f55c26f737c4778a72f1060fe963e88"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2916a42ce72648b6888c09e72c8edf5f"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2b980ecd541d4aedaebdf5a8144935e6"
}
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# 单词接龙
<p>字典 <code>wordList</code> 中从单词 <code>beginWord</code><em> </em>和 <code>endWord</code> 的 <strong>转换序列 </strong>是一个按下述规格形成的序列：</p>
<ul>
......@@ -41,33 +42,208 @@
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
int res;
string beginWord = "hit", endWord = "cog";
vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};
res = sol.ladderLength(beginWord, endWord, wordList);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> word;
for (auto s : wordList)
{
word.insert(s);
}
if (word.find(endWord) == word.end())
{
return 0;
}
queue<string> que;
que.push(beginWord);
unordered_set<string> visited;
visited.insert(beginWord);
int num = 1;
while (!que.empty())
{
int m = que.size();
for (int i = 0; i < m; i++)
{
string str = que.front();
que.pop();
if (str == endWord)
{
return num;
}
for (int j = 0; j < str.size(); j++)
{
string ss = str;
for (int k = 0; k < 26; k++)
{
char c = k + 'a';
if (c != str[j])
{
ss[j] = c;
}
if (visited.find(ss) == visited.end())
{
visited.insert(ss);
que.push(ss);
}
}
}
}
num++;
}
return 0;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> wordSet(wordList.begin(), wordList.end());
if (!wordSet.count(endWord))
return 0;
unordered_map<string, int> pathCount{{{beginWord, 1}}};
queue<string> q{{beginWord}};
while (!q.empty())
{
string word = q.front();
q.pop();
for (int i = 0; i < word.size(); i++)
{
string newWord = word;
for (char c = 'a'; c <= 'z'; c++)
{
newWord[i] = c;
if (wordSet.count(newWord) && newWord == endWord)
return pathCount[word] + 1;
if (wordSet.count(newWord) && !pathCount.count(newWord))
{
pathCount[newWord] = pathCount[word] + 1;
q.push(newWord);
}
}
}
}
return 0;
}
};
```
### B
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
queue<string> q;
map<string, int> m1;
map<string, int> re;
int n = wordList.size();
for (int i = 0; i < n; i++)
m1[wordList[i]] = 1;
re[beginWord] = 1;
q.push(beginWord);
while ((!q.empty()) && m1.size())
{
string now = q.front();
q.pop();
int num = re[now];
int llen = now.size();
for (int i = 0; i < llen; i++)
{
string temp = now;
for (char c = 'a'; c <= 'z'; c++)
{
if (temp[i] == c)
continue;
else
temp[i] = c;
if (m1.find(temp) != m1.end())
{
if (temp == endWord)
return num + 1;
q.push(temp);
re[temp] = num + 1;
m1.erase(temp);
}
}
}
}
return 0;
}
};
```
### C
```cpp
class Solution
{
public:
int ladderLength(string beginWord, string endWord, vector<string> &wordList)
{
unordered_set<string> set(wordList.begin(), wordList.end());
if (!set.count(endWord))
return 0;
queue<pair<string, int>> q;
q.push({beginWord, 1});
while (!q.empty())
{
string cur = q.front().first;
int step = q.front().second;
q.pop();
if (cur == endWord)
return step;
for (int i = 0; i < cur.size(); ++i)
{
char ch = cur[i];
for (char c = 'a'; c <= 'z'; ++c)
{
if (c == ch)
continue;
cur[i] = c;
if (set.count(cur))
{
q.push({cur, 1 + step});
set.erase(cur);
}
}
cur[i] = ch;
}
}
return 0;
}
};
```
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0655936f05744347a35896d9cbe4809a"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "08f9df5b29604656ae4f47ac37a99596"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d4afd5b7ec75416eab7ec8cc294fb520"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "a0804ac80c1246c9b44cbeb6b2564da4"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f5315a1af82e49cf975f3f2831c250de"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "e2576a710a4240ebb99b5c4942e2c389"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "caacedf5706f4cdd84999233e85715f0"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f3af026a6613431fb903713a6b305262"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "292c6821c8cf48d3a4e1bcc40a4954d1"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "1eafc7122c374b9087ecb4dfe7cdef6f"
}
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......@@ -29,33 +29,196 @@
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
int res;
vector<int> vec = {0, 3, 7, 2, 5, 8, 4, 6, 0, 1};
res = sol.longestSequence(vec);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
int longestConsecutive(vector<int> &nums)
{
unordered_set<int> hash;
for (const int &num : nums)
{
hash.insert(num);
}
int ans = 0;
while (!hash.empty())
{
int cur = *(hash.begin());
hash.erase(cur);
int next = cur + 1, prev = cur - 1;
while (hash.count(next))
{
hash.erase(next++);
}
while (hash.count(prev))
{
hash.erase(prev--);
}
ans = max(ans, next - prev);
}
return ans;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int longestConsecutive(vector<int> &nums)
{
unordered_map<int, int> mp;
int l = 0, r = 0, res = 0, len = 0;
for (int num : nums)
{
if (!mp[num])
{
l = mp[num - 1];
r = mp[num + 1];
len = l + r + 1;
res = max(res, len);
mp[num] = len;
mp[num - l] = len;
mp[num + r] = len;
}
}
return res;
}
};
```
### B
```cpp
class Solution
{
struct DisJointSet
{
vector<int> _id;
vector<int> _size;
int max_size;
int _count;
DisJointSet(int Num)
{
for (int i = 0; i < Num; i++)
{
_id.emplace_back(i);
_size.emplace_back(1);
}
_count = Num;
max_size = 1;
}
int find_(int p)
{
while (p != _id[p])
{
_id[p] = _id[_id[p]];
p = _id[p];
}
return p;
}
void _union(int p, int q)
{
int i = find_(p);
int j = find_(q);
if (i == j)
return;
if (_size[i] > _size[j])
{
_id[j] = i;
_size[i] += _size[j];
max_size = max(max_size, _size[i]);
}
else
{
_id[i] = j;
_size[j] += _size[i];
max_size = max(max_size, _size[j]);
}
_count--;
}
};
public:
int longestConsecutive(vector<int> &nums)
{
if (nums.size() == 0)
return 0;
DisJointSet disJointSet(nums.size());
unordered_set<int> nums_set;
unordered_map<int, int> nums_disJointSetID_map;
for (int i = 0; i < nums.size(); i++)
{
if (nums_set.find(nums[i]) != nums_set.end())
continue;
nums_set.insert(nums[i]);
nums_disJointSetID_map[nums[i]] = i;
if (nums_set.find(nums[i] - 1) != nums_set.end())
{
disJointSet._union(nums_disJointSetID_map[nums[i]], nums_disJointSetID_map[nums[i] - 1]);
}
if (nums_set.find(nums[i] + 1) != nums_set.end())
{
disJointSet._union(nums_disJointSetID_map[nums[i]], nums_disJointSetID_map[nums[i] + 1]);
}
}
return disJointSet.max_size;
}
};
```
### C
```cpp
class Solution
{
public:
int longestConsecutive(vector<int> &vec)
{
set<int> s;
for (int elem : vec)
s.insert(elem);
int longest = 0;
int cnt = 1;
for (auto e : s)
{
if (s.find(e + 1) != s.end())
{
cnt++;
}
else
{
cnt = 1;
}
longest = max(longest, cnt);
}
return longest;
}
};
```
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "daadb9b98a0545439c14e91a3c4c1b24"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5e8b6573c414422b96149e1426ef77f4"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "4fcba10db64b410caaf46e7382694efe"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "b6c2ccb59c3c4bbbb34157ca30736e7d"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "732941eeacb244b4b09a77736c2235b5"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "a646669108a343a29ad09ace6292d675"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fcae274ca8204cd3a0127bc7a246efb2"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "656c7aaf3a9a4501a0acf9ae97c64225"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "03d7c968fbd947b7839d55cda987f957"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "880a2e64223646969b4f5cbec063c6c2"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "436d7f604f3148719640ef4e1f259aef"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "7e15a60056114ed3823ffa14baacfc52"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "11d24e0268454dc5862b2904b218357b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "49b8393d9d9340019f988c2fb0af1a28"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "919de84d291a4d43ab328a3c82b8463a"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "e35ba70494bb4edeb14433fd1675333b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "efc83d2975444730948155122314df6c"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "a700ecd964f741f288ffa14d14363a5b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "adc8680ea7b24d70a1ffdd603056e032"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d6591d3cdc5f4e8d9d9272b849d6ea1e"
}
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......@@ -33,33 +33,249 @@
</div>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<vector<char>> board = {{'X', 'X', 'X', 'X'}, {'X', 'O', 'O', 'X'}, {'X', 'X', 'O', 'X'}, {'X', 'O', 'X', 'X'}};
sol.solve(board);
for (auto i : board)
{
for (auto j : i)
cout << j << " ";
cout << endl;
}
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
vector<vector<bool>> st;
int m, n;
void bfs(vector<vector<char>> &board, int a, int b)
{
int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
st[a][b] = true;
queue<pair<int, int>> q;
q.push(make_pair(a, b));
while (q.size())
{
pair<int, int> t = q.front();
q.pop();
for (int i = 0; i < 4; i++)
{
int x = dx[i] + t.first, y = dy[i] + t.second;
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O' && !st[x][y])
{
q.push(make_pair(x, y));
st[x][y] = true;
}
}
}
}
void solve(vector<vector<char>> &board)
{
if (!board.size())
return;
m = board.size(), n = board[0].size();
st = vector<vector<bool>>(m, vector<bool>(n, false));
for (int i = 0; i < m; i++)
{
if (board[i][0] == 'O')
bfs(board, i, 0);
if (board[i][n] == 'O')
bfs(board, i, n - 1);
}
for (int i = 0; i < n; i++)
{
if (board[0][i] == 'O')
bfs(board, 0, i);
if (board[m][i] == 'O')
bfs(board, m - 1, i);
}
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (!st[i][j])
board[i][j] = 'X';
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int n, m;
void dfs(vector<vector<char>> &board, int x, int y)
{
if (x < 0 || y < 0 || x >= m || y >= n || board[x][y] != 'O')
return;
board[x][y] = 'A';
dfs(board, x + 1, y);
dfs(board, x, y + 1);
dfs(board, x - 1, y);
dfs(board, x, y - 1);
}
void solve(vector<vector<char>> &board)
{
m = board.size();
if (m == 0)
return;
n = board[0].size();
for (int i = 0; i < m; i++)
{
dfs(board, i, 0);
dfs(board, i, n - 1);
}
for (int i = 1; i < n - 1; i++)
{
dfs(board, 0, i);
dfs(board, m - 1, i);
}
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
if (board[i][j] == 'A')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
}
};
```
### B
```cpp
class Solution
{
public:
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
queue<pair<int, int>> s;
void solve(vector<vector<char>> &board)
{
int n = board.size();
if (n <= 2)
return;
int m = board[0].size();
if (m <= 2)
return;
for (int i = 0; i < m; i++)
{
if (board[0][i] == 'O')
s.push({0, i});
if (board[n - 1][i] == 'O')
s.push({n - 1, i});
}
for (int i = 1; i < n - 1; i++)
{
if (board[i][0] == 'O')
s.push({i, 0});
if (board[i][m - 1] == 'O')
s.push({i, m - 1});
}
while (!s.empty())
{
int x = s.front().first, y = s.front().second;
board[x][y] = '#';
s.pop();
for (int i = 0; i < 4; i++)
{
int mx = x + dx[i], my = y + dy[i];
if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O')
{
continue;
}
s.push({mx, my});
}
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (board[i][j] == '#')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
}
};
```
### C
```cpp
class Solution
{
public:
void dfs(vector<vector<char>> &board, int row, int col)
{
if (row < 0 || row >= board.size() || col < 0 || col >= board[0].size())
return;
if (board[row][col] != 'O')
return;
board[row][col] = '+';
dfs(board, row + 1, col);
dfs(board, row, col + 1);
dfs(board, row - 1, col);
dfs(board, row, col - 1);
}
void solve(vector<vector<char>> &board)
{
if (board.size() <= 1 || board[0].size() <= 2)
return;
for (int i = 0; i < board.size(); i++)
{
if (board[i][0] == 'O')
dfs(board, i, 0);
if (board[i][board[0].size() - 1] == 'O')
dfs(board, i, board[0].size() - 1);
}
for (int i = 0; i < board[0].size(); i++)
{
if (board[0][i] == 'O')
dfs(board, 0, i);
if (board[board.size() - 1][i] == 'O')
dfs(board, board.size() - 1, i);
}
for (int i = 0; i < board.size(); i++)
{
for (int j = 0; j < board[0].size(); j++)
{
if (board[i][j] == '+')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
}
};
```
data_backup/1.leetcode/1300-转变数组后最接近目标值的数组和/solution.json 0 → 100644
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{
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"author": "CSDN.net",
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"author": "CSDN.net",
"source": "solution.md",
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
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}
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6ff80de10e964eab90b3a79e51bd5e3b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d160b289d601480d9196307cf9bc3af4"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "11317f6e2b3d435595ce32558fe30fb7"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6fb4b2c92ee8414f881f759ab8520572"
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"exercise_id": "9123c1a844c74efea4702e102c31548a"
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{
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"author": "CSDN.net",
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"exercise_id": "e331d492c24b479383778a63c5f9a2c7"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "d3edd0edca8440578e30a448a8af2f63"
}
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "ed2ee7c0f37042d5b4adf16663b9f4b7"
}
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2daf9dfb1e6e4ac1b12a7d11d64600c9"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2951640ac01646369cefeb4c48903905"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "28486d69b0eb42e1acec7c55ec43d5ee"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "bd2665b70aa448068bba3399b95b11ed"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "1fbf4f117392439aac363595cdef6fa4"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "251e6f305b1f45c19fc9dc2827943bee"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "39932d0226654481890d2a008648a8f5"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "9b812dcf701c4c57b4302f17354e5176"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "ba08167d4b2b49a6a55afab018358382"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "be2c2f901f4347ab8b3969c58cdbc74f"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0bc439ea08d24a7db3013a0a6dc5fc6b"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5c3440d4aa364533be444e4452f284b5"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "a8c2acfed25f4d91b39e892ae27449e3"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "71622c8ce7184e94a6250bed4850957a"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0eb57565424d430999469b071d71d7bd"
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"type": "code_options",
"author": "CSDN.net",
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"exercise_id": "4c5113e74c4749fea37b80f3acb43be2"
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"author": "CSDN.net",
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"exercise_id": "a7248577d79348c0b392223b58b9305d"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0477ca8c5c4948559486c59e6f0ee6fa"
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "e2c2b6032d2d441a9936476bb3c1a6c5"
}
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"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0028da9e903d44f2b9b46dfeffca5458"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5cd1f37dfa0a4ada847eedcc222d127b"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0209374b0d234f289e8b5d4930be5788"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "725ff052dc5b4f21ad2cca00898125e7"
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{
"type": "code_options",
"author": "CSDN.net",
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{
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"exercise_id": "441e6b155c9c48b086041fb1cf14e5c7"
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{
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"source": "solution.md",
"exercise_id": "5153d58b7d624ce9873e531da1aebdcc"
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{
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"author": "CSDN.net",
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{
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"author": "CSDN.net",
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"exercise_id": "f46872c2e0e24314a0cb9a7866a8a882"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "aae14ec3b4154df09d4c3c85a0d299f7"
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"source": "solution.md",
"exercise_id": "155f4930775f4ceab42c8551077b2050"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "e24c0f5e13c94c7f830911eacd3072c3"
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fe9fd61ba9f342438c35b5ff012045d5"
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "21d33101880d470685d85b0d0330103a"
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"source": "solution.md",
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"source": "solution.md",
"exercise_id": "df2b140bc3f6423187cee1d96cbda2ee"
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "3b0b415c69e940159ccd5a0738244c59"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "7175a1cb92a142418583d4ff437923e6"
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......@@ -54,33 +54,213 @@
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
string res;
int numerator = 1, denominator = 2;
res = sol.fractionToDecimal(numerator, denominator);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
const static int CACHE_SIZE = 1000;
string fractionToDecimal(int numerator, int denominator)
{
if (numerator == 0)
return "0";
bool isNegative = numerator < 0 ^ denominator < 0;
double doubleNumerator = fabs(numerator);
double doubleDenominator = fabs(denominator);
double integral = floor(doubleNumerator / doubleDenominator);
doubleNumerator -= doubleDenominator * integral;
int len = 0;
string cache;
vector<double> dividendCache;
bool isRecurring = false;
int recurringStart = 0;
int i = 0;
while (doubleNumerator)
{
doubleNumerator *= 10;
len = dividendCache.size();
for (i = 0; i < len; ++i)
if (dividendCache[i] == doubleNumerator)
{
isRecurring = true;
recurringStart = i;
break;
}
if (isRecurring)
break;
i = (int)(floor(doubleNumerator / doubleDenominator));
cache += i;
dividendCache.push_back(doubleNumerator);
doubleNumerator -= doubleDenominator * i;
}
string ret = isNegative ? "-" : "";
string tmp;
char c;
if (integral == 0)
ret += "0";
else
{
while (integral)
{
c = (int)(integral - 10 * floor(integral / 10)) + '0';
tmp = c + tmp;
integral = floor(integral / 10);
}
ret += tmp;
}
if (dividendCache.size() > 0)
ret += '.';
i = 0;
if (isRecurring)
{
ret += cache.substr(0, recurringStart);
ret += '(';
ret += cache.substr(recurringStart);
ret += ')';
}
else
ret += cache;
return ret;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
typedef long long LL;
LL x = numerator, y = denominator;
if (x % y == 0)
return to_string(x / y);
string res;
if ((x < 0) ^ (y < 0))
res += '-';
x = abs(x), y = abs(y);
res += to_string(x / y) + '.';
x %= y;
unordered_map<LL, int> hash;
while (x)
{
hash[x] = res.size();
x *= 10;
res += to_string(x / y);
x %= y;
if (hash.count(x))
{
res = res.substr(0, hash[x]) + '(' + res.substr(hash[x]) + ')';
break;
}
}
return res;
}
};
```
### B
```cpp
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
string ans;
unordered_map<long, int> use;
if (numerator > 0 && denominator < 0 || numerator < 0 && denominator > 0)
ans = '-';
long numerat = fabs(numerator);
long denominat = fabs(denominator);
ans += to_string(numerat / denominat);
numerat = numerat % denominat;
int point, x;
if (numerat)
{
ans += ".";
point = ans.length() - 1;
}
else
return ans;
while (numerat && use.count(numerat) == 0)
{
use[numerat] = ++point;
numerat *= 10;
x = numerat / denominat;
numerat = numerat % denominat;
ans.push_back(x + '0');
}
if (numerat)
{
ans.insert(ans.begin() + use[numerat], '(');
ans.push_back(')');
}
return ans;
}
};
```
### C
```cpp
class Solution
{
public:
string fractionToDecimal(int numerator, int denominator)
{
int m1 = numerator > 0 ? 1 : -1;
int m2 = denominator > 0 ? 1 : -1;
long long num = abs((long long)numerator);
long long den = abs((long long)denominator);
long long t = num / den;
long long res = num % den;
string ans = to_string(t);
if (m1 * m2 == -1 && (t > 0 || res > 0))
ans = '-' + ans;
if (res == 0)
return ans;
ans += '.';
unordered_map<long long, int> m;
string s = "";
int pos = 0;
while (res != 0)
{
if (m.find(res) != m.end())
{
s.insert(m[res], "(");
s += ')';
return ans + s;
}
m[res] = pos;
s += to_string((res * 10) / den);
res = (res * 10) % den;
pos++;
}
return ans + s;
}
};
```
data_backup/1.leetcode/1660-纠正二叉树/solution.json 0 → 100644
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{
"type": "code_options",
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2ef6049026cf4dc3b4c12551fda8fe5c"
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{
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"author": "CSDN.net",
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"author": "CSDN.net",
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"exercise_id": "f82dfdf0b4704fdeaa0989cd7f2e7fa0"
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{
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "04c31d33b27344cfb96bf1b2179b6798"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "ec7bbd853fda427f82b06d9a006bf406"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "b3645dc3cd034093bede4f47df5f25cf"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "7c76ecf38c0640cc92d5f33e6ac5e406"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "cc10cccca673444cb35c834ce09e5296"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "38295a128d5e404fa7fdf0c05c8b6469"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f1b95c135c5d4ab1ac4c5cf73741c907"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "6c8d358e60754761a568b8a2a22ee0f8"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "e61b6b4cf9584caa960bf1906c388c95"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f4897899fcb74526b14cc76b0e1eeca6"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "012b29fa9d764d908d42e16cfc803993"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "16a1527854e74455a5284d347e8dfc26"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "18f41bad33934fe6aa0c2b16c7ae5c16"
}
\ No newline at end of file
data_backup/1.leetcode/1678-设计 Goal 解析器/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "53b49f9b686d441eb755bad68ff3ffc0"
}
\ No newline at end of file
data_backup/1.leetcode/1679-K 和数对的最大数目/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fcde4897474849d4ab629be1797b8aca"
}
\ No newline at end of file
data_backup/1.leetcode/1680-连接连续二进制数字/solution.json 0 → 100644
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2eebc072f17e49878b6a59251eb0c52c"
}
\ No newline at end of file
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "b67f602a50b349db8fb21a73afca41e0"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "2c98d44a52a24e3db4adca70ce51f7f2"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "70e0f7713a84410da8795d0f105a911b"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0a3322c57b764ff486adf31cba87259e"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "5711c801f88742d7a8f5d678f0a3ba92"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "0cf9f67fbe4b4f6bab0248ad4dfdafa6"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "01a8f73bdefb4a25a6a0dfc743555685"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "605f6a8437014380b3a3664ee6ed146c"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "28f04f298f40476b8687cf64fe4accfd"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "a7faa961fc454505bf3d14798e550eec"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "837ee3d12dc7461ca1c70fe761c44ff2"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "b7ee0a3b44ee4c7396bac45eb201104e"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "a76101bb61704e2ba01b8a2ce372ea11"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "adda6958148c4598b7cd0577952e8246"
}
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "f2bb4e760d094e8f85d8b2ed13484138"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "9666fd8c55b1466399506b088f7218eb"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "be176939f89847b1a5ed9560dfa7bb96"
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
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{
"type": "code_options",
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"source": "solution.md",
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "fd243acd14e3480ab5267191e077004e"
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{
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{
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"source": "solution.md",
"exercise_id": "81c83fb16613489a843276b984c683d3"
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{
"type": "code_options",
"author": "CSDN.net",
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{
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{
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{
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{
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{
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{
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{
"type": "code_options",
"author": "CSDN.net",
"source": "solution.md",
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{
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{
"type": "code_options",
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"source": "solution.md",
"exercise_id": "0950f354fb254e94af535def2a1ba8cf"
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{
"type": "code_options",
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"exercise_id": "1d461773030941478787fed0e4f9738d"
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"author": "CSDN.net",
"source": "solution.md",
"exercise_id": "bb2d9083b13c474fa267dc044c86ab74"
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{
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{
"type": "code_options",
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"source": "solution.md",
"exercise_id": "59d8d6685a704560a093feece2137b84"
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{
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......@@ -57,33 +57,197 @@ bSTIterator.hasNext(); // 返回 False
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```
### after
```cpp
/**
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator* obj = new BSTIterator(root);
* int param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/
```
## 答案
```cpp
class BSTIterator
{
private:
void traverse(TreeNode *root, vector<int> &res)
{
if (root == nullptr)
{
return;
}
traverse(root->right, res);
traverse(root->left, res);
res.push_back(root->val);
}
vector<int> inorder(TreeNode *root)
{
vector<int> res;
traverse(root, res);
return res;
}
vector<int> arr;
int idx;
public:
BSTIterator(TreeNode *root) : idx(0), arr(inorder(root))
{
}
int next()
{
return arr[idx++];
}
bool hasNext()
{
return idx < arr.size();
}
};
```
## 选项
### A
```cpp
class BSTIterator
{
private:
void inorder(TreeNode *root, vector<int> &vec)
{
if (!root)
return;
inorder(root->left, vec);
vec.emplace_back(root->val);
inorder(root->right, vec);
}
vector<int> inorderTraversal(TreeNode *root)
{
vector<int> ret;
inorder(root, ret);
return ret;
}
vector<int> buf;
size_t idx;
public:
BSTIterator(TreeNode *root)
{
idx = 0;
buf = inorderTraversal(root);
}
int next()
{
return buf[idx++];
}
bool hasNext()
{
return idx < buf.size();
}
};
```
### B
```cpp
class BSTIterator
{
public:
stack<TreeNode *> treeMin;
BSTIterator(TreeNode *root)
{
TreeNode *t = root;
while (t)
{
treeMin.push(t);
t = t->left;
}
}
/** @return the next smallest number */
int next()
{
TreeNode *tmp = treeMin.top();
int res = tmp->val;
treeMin.pop();
tmp = tmp->right;
while (tmp)
{
treeMin.push(tmp);
tmp = tmp->left;
}
return res;
}
/** @return whether we have a next smallest number */
bool hasNext()
{
if (treeMin.empty())
return false;
else
return true;
}
};
```
### C
```cpp
class BSTIterator
{
public:
queue<int> q;
BSTIterator(TreeNode *root)
{
inorder(root, q);
}
void inorder(TreeNode *root, queue<int> &q)
{
if (root == NULL)
return;
inorder(root->left, q);
q.push(root->val);
inorder(root->right, q);
}
/** @return the next smallest number */
int next()
{
int tmp = q.front();
q.pop();
return tmp;
}
/** @return whether we have a next smallest number */
bool hasNext()
{
if (q.empty())
return false;
else
return true;
}
};
```
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# 最大数
<p>给定一组非负整数 <code>nums</code>，重新排列每个数的顺序（每个数不可拆分）使之组成一个最大的整数。</p>
<p><strong>注意：</strong>输出结果可能非常大，所以你需要返回一个字符串而不是整数。</p>
......@@ -42,33 +43,120 @@
</ul>
<p>以下错误的选项是？</p>
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
vector<int> nums = {3, 30, 34, 5, 9};
string res = sol.largestNumber(nums);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
string ans;
for (int i = 0; i < nums.size(); i++)
{
for (int j = nums.size() - 1; j > i; j--)
{
if (to_string(nums[j - 1]) + to_string(nums[j]) > to_string(nums[j]) + to_string(nums[j - 1]))
{
int tmp = nums[j - 1];
nums[j - 1] = nums[j];
nums[j] = tmp;
}
}
ans += to_string(nums[i]);
}
if (ans + "0" == "0" + ans)
return "0";
return ans;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
string ans;
sort(nums.begin(), nums.end(), [](int a, int b)
{ return to_string(a) + to_string(b) > to_string(b) + to_string(a); });
for (auto it : nums)
{
ans += to_string(it);
}
return ans[0] == '0' ? "0" : ans;
}
};
```
### B
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
if (*max_element(nums.begin(), nums.end()) == 0)
return to_string(0);
vector<string> vec(nums.size());
for (int i = 0; i < nums.size(); ++i)
vec[i] = to_string(nums[i]);
sort(vec.begin(), vec.end(), [](const string &s1, const string &s2)
{ return s1 + s2 > s2 + s1; });
return accumulate(vec.begin(), vec.end(), string());
}
};
```
### C
```cpp
class Solution
{
public:
string largestNumber(vector<int> &nums)
{
string res = "";
int n = nums.size();
vector<string> tmp;
for (int i = 0; i < n; i++)
{
tmp.push_back(to_string(nums[i]));
}
sort(tmp.begin(), tmp.end(), compare);
for (int i = 0; i < tmp.size(); i++)
{
res += tmp[i];
}
if ('0' == res[0])
{
return "0";
}
else
{
return res;
}
}
};
```
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data_backup/1.leetcode/1899-合并若干三元组以形成目标三元组/solution.json 0 → 100644
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{
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data_backup/1.leetcode/1900-最佳运动员的比拼回合/solution.json 0 → 100644
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{
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data_backup/1.leetcode/1901-找出顶峰元素 II/solution.json 0 → 100644
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{
"type": "code_options",
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data_backup/1.leetcode/1902-Depth of BST Given Insertion Order/solution.json 0 → 100644
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{
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