diff --git "a/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/config.json" "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/config.json" new file mode 100644 index 0000000000000000000000000000000000000000..e7ad92dc5e65fe4292e0ad21e308d7297580a6da --- /dev/null +++ "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/config.json" @@ -0,0 +1,12 @@ +{ + "node_id": "569d5e11c4fc5de7844053d9a733c5e8", + "keywords": [ + "leetcode", + "交错字符串" + ], + "children": [], + "export": [ + "solution.json" + ], + "title": "交错字符串" +} \ No newline at end of file diff --git "a/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/desc.html" "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/desc.html" new file mode 100644 index 0000000000000000000000000000000000000000..7dfd6b6e48ca95ca0d630790ec3cdae1237ef7be --- /dev/null +++ "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/desc.html" @@ -0,0 +1 @@ +

给定三个字符串 s1、s2、s3，请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。

两个字符串 s 和 t 交错 的定义与过程如下，其中每个字符串都会被分割成若干 非空 子字符串：

提示：a + b 意味着字符串 a 和 b 连接。

 

示例 1：

输入：s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出：true

示例 2：

输入：s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出：false

示例 3：

输入：s1 = "", s2 = "", s3 = ""
输出：true

 

提示：

\ No newline at end of file diff --git "a/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.cpp" "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.cpp" new file mode 100644 index 0000000000000000000000000000000000000000..9c77331645ffa652d66bec428913fb53748872df --- /dev/null +++ "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.cpp" @@ -0,0 +1,50 @@ +#include +#include +#include +#include +static bool isInterleave(char *s1, char *s2, char *s3) +{ + int i, j; + int len1 = strlen(s1); + int len2 = strlen(s2); + int len3 = strlen(s3); + if (len1 + len2 != len3) + { + return false; + } + bool *table = malloc((len1 + 1) * (len2 + 1) * sizeof(bool)); + bool **dp = malloc((len1 + 1) * sizeof(bool *)); + for (i = 0; i < len1 + 1; i++) + { + dp[i] = &table[i * (len2 + 1)]; + } + dp[0][0] = true; + for (i = 1; i < len1 + 1; i++) + { + dp[i][0] = dp[i - 1][0] && s1[i - 1] == s3[i - 1]; + } + for (i = 1; i < len2 + 1; i++) + { + dp[0][i] = dp[0][i - 1] && s2[i - 1] == s3[i - 1]; + } + for (i = 1; i < len1 + 1; i++) + { + for (j = 1; j < len2 + 1; j++) + { + bool up = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]; + bool left = dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]; + dp[i][j] = up || left; + } + } + return dp[len1][len2]; +} +int main(int argc, char **argv) +{ + if (argc != 4) + { + fprintf(stderr, "Usage: ./test s1 s2 s3\n"); + exit(-1); + } + printf("%s\n", isInterleave(argv[1], argv[2], argv[3]) ? "true" : "false"); + return 0; +} \ No newline at end of file diff --git "a/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.json" "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.json" new file mode 100644 index 0000000000000000000000000000000000000000..1ab33572acd40fdf0d5f2d89a84fff8059a0a5d4 --- /dev/null +++ "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.json" @@ -0,0 +1,6 @@ +{ + "type": "code_options", + "author": "CSDN.net", + "source": "solution.md", + "exercise_id": "c44af29fd6514841a9306e625c0e8a14" +} \ No newline at end of file diff --git "a/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.md" "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.md" new file mode 100644 index 0000000000000000000000000000000000000000..c6bf37221b875b9c0c64306c012cfd97adc0615f --- /dev/null +++ "b/data/2.\347\256\227\346\263\225\344\270\255\351\230\266/3.leetcode-\345\255\227\347\254\246\344\270\262/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.md" @@ -0,0 +1,158 @@ +# 交错字符串 +

给定三个字符串 s1、s2、s3，请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。

两个字符串 s 和 t 交错 的定义与过程如下，其中每个字符串都会被分割成若干 非空 子字符串：

  • s = s1 + s2 + ... + sn
  • t = t1 + t2 + ... + tm
  • |n - m| <= 1
  • 交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...

提示：a + b 意味着字符串 a 和 b 连接。

 

示例 1：

输入：s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出：true

示例 2：

输入：s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出：false

示例 3：

输入：s1 = "", s2 = "", s3 = ""
输出：true

 

提示：

  • 0 <= s1.length, s2.length <= 100
  • 0 <= s3.length <= 200
  • s1、s2、和 s3 都由小写英文字母组成
+

以下错误的选项是？

+## aop +### before +```cpp +#include +using namespace std; +``` +### after +```cpp +int main() +{ + Solution sol; + bool res; + string s1 = "aabcc"; + string s2 = "dbbca"; + string s3 = "aadbbcbcac"; + res = sol.isInterleave(s1, s2, s3); + cout << res; + return 0; +} +``` + +## 答案 +```cpp +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + int len1 = s1.length(); + int len2 = s2.length(); + int len3 = s3.length(); + + if (len1 + len2 != len3) + return false; + + bool f[len1 + 1][len2 + 1]; + f[0][0] = true; + for (int i = 0; i < len1 + 1; i++) + { + for (int j = 0; j < len2 + 1; j++) + { + + if (j > 0) + { + f[i][j] = f[i][j - 1] && s3[i + j - 1]; + } + if (i > 0) + { + f[i][j] = f[i][j] || (f[i - 1][j] && s3[i + j - 1]); + } + } + } + return f[len1][len2]; + } +}; +``` +## 选项 + +### A +```cpp +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + if (s1.size() + s2.size() != s3.size()) + return false; + int m = s1.size(), n = s2.size(), i, j, k; + vector> dp(m + 1, vector(n + 1, 0)); + + dp[0][0] = 1; + for (i = 0; i < m; i++) + if (s1[i] == s3[i]) + dp[i + 1][0] = 1; + else + break; + for (i = 0; i < n; i++) + if (s2[i] == s3[i]) + dp[0][i + 1] = 1; + else + break; + + for (i = 1; i <= m; ++i) + for (j = 1; j <= n; j++) + { + k = i + j; + if (s1[i - 1] == s3[k - 1]) + dp[i][j] |= dp[i - 1][j]; + if (s2[j - 1] == s3[k - 1]) + dp[i][j] |= dp[i][j - 1]; + } + return dp[m][n]; + } +}; +``` + +### B +```cpp +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + if (s1.size() + s2.size() != s3.size()) + return false; + int m = s1.size(), n = s2.size(); + vector> dp(m + 1, vector(n + 1, false)); + dp[0][0] = true; + for (int i = 1; i <= m; ++i) + dp[i][0] = dp[i - 1][0] && (s1[i - 1] == s3[i - 1]); + for (int i = 1; i <= n; ++i) + dp[0][i] = dp[0][i - 1] && (s2[i - 1] == s3[i - 1]); + for (int i = 1; i <= m; ++i) + for (int j = 1; j <= n; ++j) + { + dp[i][j] = (dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]) || (dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]); + } + return dp[m][n]; + } +}; +``` + +### C +```cpp +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + int s1_len = s1.size(); + int s2_len = s2.size(); + int s3_len = s3.size(); + if (s1_len + s2_len != s3_len) + return false; + if (s1_len == 0 || s2_len == 0) + return s1 + s2 == s3; + vector> dp(s1_len + 1, vector(s2_len + 1)); + dp[0][0] = true; + for (int i = 0; i <= s1_len; i++) + { + for (int j = 0; j <= s2_len; j++) + { + if (dp[i][j]) + { + if (i < s1_len && s1[i] == s3[i + j]) + dp[i + 1][j] = true; + if (j < s2_len && s2[j] == s3[i + j]) + dp[i][j + 1] = true; + } + } + } + return dp[s1_len][s2_len]; + } +}; +``` diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/config.json" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/config.json" new file mode 100644 index 0000000000000000000000000000000000000000..1e7348c90e3ee1c5d5f8302e35bce191c43e2225 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/config.json" @@ -0,0 +1,12 @@ +{ + "node_id": "569d5e11c4fc5de7844053d9a733c5e8", + "keywords": [ + "leetcode", + "恢复二叉搜索树" + ], + "children": [], + "export": [ + "solution.json" + ], + "title": "恢复二叉搜索树" +} \ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/desc.html" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/desc.html" new file mode 100644 index 0000000000000000000000000000000000000000..a4c049d29c48dfe44b2a5c28e856712880ea4353 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/desc.html" @@ -0,0 +1 @@ +

给你二叉搜索树的根节点 root ，该树中的两个节点被错误地交换。请在不改变其结构的情况下，恢复这棵树。

进阶：使用 O(n) 空间复杂度的解法很容易实现。你能想出一个只使用常数空间的解决方案吗？

 

示例 1：

输入：root = [1,3,null,null,2]
输出：[3,1,null,null,2]
解释：3 不能是 1 左孩子，因为 3 > 1 。交换 1 和 3 使二叉搜索树有效。

示例 2：

输入：root = [3,1,4,null,null,2]
输出：[2,1,4,null,null,3]
解释：2 不能在 3 的右子树中，因为 2 < 3 。交换 2 和 3 使二叉搜索树有效。

 

提示：

  • 树上节点的数目在范围 [2, 1000] 内
  • -231 <= Node.val <= 231 - 1
\ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.cpp" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.cpp" new file mode 100644 index 0000000000000000000000000000000000000000..9a885cee574a7664ea0fc3b4604f332ddc5d38c0 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.cpp" @@ -0,0 +1,49 @@ +#include +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) {} +}; +class Solution +{ +public: + void recoverTree(TreeNode *root) + { + dfs(root); + int tmp = p0_->val; + p0_->val = p1_->val; + p1_->val = tmp; + } +private: + int wrong_ = 0; + TreeNode *prev_ = nullptr; + TreeNode *p0_ = nullptr; + TreeNode *p1_ = nullptr; + void dfs(TreeNode *root) + { + if (root == nullptr || wrong_ == 2) + { + return; + } + dfs(root->left); + if (prev_ != nullptr && prev_->val > root->val) + { + if (++wrong_ == 1) + { + p0_ = prev_; + p1_ = root; + } + else if (wrong_ == 2) + { + p1_ = root; + } + } + prev_ = root; + dfs(root->right); + } +}; \ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.json" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.json" new file mode 100644 index 0000000000000000000000000000000000000000..c2dc9be962f6bc732180aa682dbee1b0154f9f76 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.json" @@ -0,0 +1,6 @@ +{ + "type": "code_options", + "author": "CSDN.net", + "source": "solution.md", + "exercise_id": "6f36dd5736a941d091ed802411238645" +} \ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.md" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.md" new file mode 100644 index 0000000000000000000000000000000000000000..a6c4158b758a9622e727312bb9c5df7c43af088a --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.md" @@ -0,0 +1,190 @@ +# 恢复二叉搜索树 +

给你二叉搜索树的根节点 root ，该树中的两个节点被错误地交换。请在不改变其结构的情况下，恢复这棵树。

进阶：使用 O(n) 空间复杂度的解法很容易实现。你能想出一个只使用常数空间的解决方案吗？

 

示例 1：

输入：root = [1,3,null,null,2]
输出：[3,1,null,null,2]
解释：3 不能是 1 左孩子，因为 3 > 1 。交换 1 和 3 使二叉搜索树有效。

示例 2：

输入：root = [3,1,4,null,null,2]
输出：[2,1,4,null,null,3]
解释：2 不能在 3 的右子树中，因为 2 < 3 。交换 2 和 3 使二叉搜索树有效。

 

提示：

  • 树上节点的数目在范围 [2, 1000] 内
  • -231 <= Node.val <= 231 - 1
+

以下错误的选项是？

+## aop +### before +```cpp +#include +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: + TreeNode *pre = NULL; + TreeNode *one = NULL; + TreeNode *two = NULL; + bool search(TreeNode *root) + { + if (root == NULL) + return false; + if (search(root->left)) + return true; + if (pre != NULL && pre->val > root->val) + { + if (one == NULL) + { + one = root; + two = pre; + } + else + { + two = root; + return true; + } + } + pre = root; + if (search(root->right)) + return true; + return false; + } + void recoverTree(TreeNode *root) + { + search(root); + swap(one->val, two->val); + } +}; +``` +## 选项 + +### A +```cpp +class Solution +{ +public: + TreeNode *x, *y, *pre; + void DFS(TreeNode *root) + { + if (root == nullptr) + return; + DFS(root->left); + if (pre && root->val < pre->val) + { + x = root; + if (!y) + y = pre; + else + return; + } + pre = root; + DFS(root->right); + } + void recoverTree(TreeNode *root) + { + DFS(root); + swap(x->val, y->val); + } +}; +``` + +### B +```cpp +class Solution +{ +public: + void insert(int val, vector &vals) + { + if (vals.size() > 0) + { + for (int i = 0; i < vals.size(); i++) + { + if (val < vals[i]) + { + vals.insert(vals.begin() + i, val); + return; + } + } + } + vals.push_back(val); + } + void recoverTree(TreeNode *root) + { + stack s; + vector vals; + TreeNode *tem = root; + while (tem || !s.empty()) + { + while (tem) + { + s.push(tem); + tem = tem->left; + } + if (!s.empty()) + { + tem = s.top(); + s.pop(); + insert(tem->val, vals); + tem = tem->right; + } + } + tem = root; + int j = 0; + while (tem || !s.empty()) + { + while (tem) + { + s.push(tem); + tem = tem->left; + } + if (!s.empty()) + { + tem = s.top(); + s.pop(); + tem->val = vals[j++]; + tem = tem->right; + } + } + } +}; +``` + +### C +```cpp +class Solution +{ +public: + vector order; + void inOrder(TreeNode *root) + { + if (root->left) + inOrder(root->left); + order.push_back(root); + if (root->right) + inOrder(root->right); + } + void recoverTree(TreeNode *root) + { + inOrder(root); + int left = -1, right = -1; + for (int i = 0; i < order.size() - 1; ++i) + { + if (order[i]->val > order[i + 1]->val) + { + if (left == -1) + left = i; + right = i + 1; + } + } + if (right == -1) + swap(order[left]->val, order[left + 1]->val); + else + swap(order[left]->val, order[right]->val); + } +}; +``` diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/config.json" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/config.json" new file mode 100644 index 0000000000000000000000000000000000000000..36f471b825d9ffff868d4375868fa44e807f3b25 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/config.json" @@ -0,0 +1,12 @@ +{ + "node_id": "569d5e11c4fc5de7844053d9a733c5e8", + "keywords": [ + "leetcode", + "相同的树" + ], + "children": [], + "export": [ + "solution.json" + ], + "title": "相同的树" +} \ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/desc.html" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/desc.html" new file mode 100644 index 0000000000000000000000000000000000000000..2d8a4c0029cd3168cf01d57a2c8e65b6c71649ae --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/desc.html" @@ -0,0 +1 @@ +

给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。

如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。

 

示例 1：

输入：p = [1,2,3], q = [1,2,3]
输出：true

示例 2：

输入：p = [1,2], q = [1,null,2]
输出：false

示例 3：

输入：p = [1,2,1], q = [1,1,2]
输出：false

 

提示：

  • 两棵树上的节点数目都在范围 [0, 100] 内
  • -104 <= Node.val <= 104
\ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.cpp" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.cpp" new file mode 100644 index 0000000000000000000000000000000000000000..d465336018a8577aa7e9e889e1e3f8876d0eb484 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.cpp" @@ -0,0 +1,31 @@ +#include +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) {} +}; +class Solution +{ +public: + bool isSameTree(TreeNode *p, TreeNode *q) + { + if (p == nullptr && q == nullptr) + { + return true; + } + if (p == nullptr || q == nullptr) + { + return false; + } + if (p->val != q->val) + { + return false; + } + return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); + } +}; \ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.json" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.json" new file mode 100644 index 0000000000000000000000000000000000000000..c6382daa809df6a4f4a1042630cb5229b44496f0 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.json" @@ -0,0 +1,6 @@ +{ + "type": "code_options", + "author": "CSDN.net", + "source": "solution.md", + "exercise_id": "0f00b8981cd541598110de5d1445d8f6" +} \ No newline at end of file diff --git "a/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.md" "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.md" new file mode 100644 index 0000000000000000000000000000000000000000..c58f0f3ecb4ff5179e5bf047974b64c14be21eb9 --- /dev/null +++ "b/data/3.\347\256\227\346\263\225\351\253\230\351\230\266/5.leetcode-\346\240\221/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.md" @@ -0,0 +1,90 @@ +# 相同的树 +

给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。

如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。

 

示例 1：

输入：p = [1,2,3], q = [1,2,3]
输出：true

示例 2：

输入：p = [1,2], q = [1,null,2]
输出：false

示例 3：

输入：p = [1,2,1], q = [1,1,2]
输出：false

 

提示：

  • 两棵树上的节点数目都在范围 [0, 100] 内
  • -104 <= Node.val <= 104
+

以下错误的选项是？

+## aop +### before +```cpp +#include +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 +都是错的 +``` +## 选项 + +### A +```cpp +class Solution +{ +public: + bool isSameTree(TreeNode *p, TreeNode *q) + { + if (p == nullptr) + return q == nullptr; + return q != nullptr && p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right); + } +}; +``` + +### B +```cpp +class Solution +{ +public: + bool isSameTree(TreeNode *p, TreeNode *q) + { + TreeNode *temp, *temq; + temp = p; + temq = q; + bool lb, rb; + if (!p && !q) + return true; + if ((!p || !q) || p->val != q->val) + return false; + else + { + lb = isSameTree(p->left, q->left); + rb = isSameTree(p->right, q->right); + } + if (lb && rb) + return true; + else + return false; + } +}; +``` + +### C +```cpp +class Solution +{ +public: + bool isSameTree(TreeNode *p, TreeNode *q) + { + if (p == NULL && q == NULL) + return true; + if (p == NULL || q == NULL) + return false; + if (p->val != q->val) + return false; + return (isSameTree(p->left, q->left) && isSameTree(p->right, q->right)); + } +}; +``` diff --git "a/data_backup/1.leetcode/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.md" "b/data_backup/1.leetcode/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.md" index 5a075ca4c2e2379de9f7ebfcbf55be1d71d0f1be..c6bf37221b875b9c0c64306c012cfd97adc0615f 100644 --- "a/data_backup/1.leetcode/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.md" +++ "b/data_backup/1.leetcode/96_\344\272\244\351\224\231\345\255\227\347\254\246\344\270\262/solution.md" @@ -4,30 +4,155 @@ ## aop ### before ```cpp - +#include +using namespace std; ``` ### after ```cpp - +int main() +{ + Solution sol; + bool res; + string s1 = "aabcc"; + string s2 = "dbbca"; + string s3 = "aadbbcbcac"; + res = sol.isInterleave(s1, s2, s3); + cout << res; + return 0; +} ``` ## 答案 ```cpp +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + int len1 = s1.length(); + int len2 = s2.length(); + int len3 = s3.length(); + + if (len1 + len2 != len3) + return false; + + bool f[len1 + 1][len2 + 1]; + f[0][0] = true; + for (int i = 0; i < len1 + 1; i++) + { + for (int j = 0; j < len2 + 1; j++) + { + if (j > 0) + { + f[i][j] = f[i][j - 1] && s3[i + j - 1]; + } + if (i > 0) + { + f[i][j] = f[i][j] || (f[i - 1][j] && s3[i + j - 1]); + } + } + } + return f[len1][len2]; + } +}; ``` ## 选项 ### A ```cpp +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + if (s1.size() + s2.size() != s3.size()) + return false; + int m = s1.size(), n = s2.size(), i, j, k; + vector> dp(m + 1, vector(n + 1, 0)); + dp[0][0] = 1; + for (i = 0; i < m; i++) + if (s1[i] == s3[i]) + dp[i + 1][0] = 1; + else + break; + for (i = 0; i < n; i++) + if (s2[i] == s3[i]) + dp[0][i + 1] = 1; + else + break; + + for (i = 1; i <= m; ++i) + for (j = 1; j <= n; j++) + { + k = i + j; + if (s1[i - 1] == s3[k - 1]) + dp[i][j] |= dp[i - 1][j]; + if (s2[j - 1] == s3[k - 1]) + dp[i][j] |= dp[i][j - 1]; + } + return dp[m][n]; + } +}; ``` ### B ```cpp - +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + if (s1.size() + s2.size() != s3.size()) + return false; + int m = s1.size(), n = s2.size(); + vector> dp(m + 1, vector(n + 1, false)); + dp[0][0] = true; + for (int i = 1; i <= m; ++i) + dp[i][0] = dp[i - 1][0] && (s1[i - 1] == s3[i - 1]); + for (int i = 1; i <= n; ++i) + dp[0][i] = dp[0][i - 1] && (s2[i - 1] == s3[i - 1]); + for (int i = 1; i <= m; ++i) + for (int j = 1; j <= n; ++j) + { + dp[i][j] = (dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]) || (dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]); + } + return dp[m][n]; + } +}; ``` ### C ```cpp - +class Solution +{ +public: + bool isInterleave(string s1, string s2, string s3) + { + int s1_len = s1.size(); + int s2_len = s2.size(); + int s3_len = s3.size(); + if (s1_len + s2_len != s3_len) + return false; + if (s1_len == 0 || s2_len == 0) + return s1 + s2 == s3; + vector> dp(s1_len + 1, vector(s2_len + 1)); + dp[0][0] = true; + for (int i = 0; i <= s1_len; i++) + { + for (int j = 0; j <= s2_len; j++) + { + if (dp[i][j]) + { + if (i < s1_len && s1[i] == s3[i + j]) + dp[i + 1][j] = true; + if (j < s2_len && s2[j] == s3[i + j]) + dp[i][j + 1] = true; + } + } + } + return dp[s1_len][s2_len]; + } +}; ``` diff --git "a/data_backup/1.leetcode/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.md" "b/data_backup/1.leetcode/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.md" index 02f150cc85387262bf45ef586514a9d56e8d3aca..a6c4158b758a9622e727312bb9c5df7c43af088a 100644 --- "a/data_backup/1.leetcode/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.md" +++ "b/data_backup/1.leetcode/98_\346\201\242\345\244\215\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221/solution.md" @@ -4,7 +4,18 @@ ## aop ### before ```cpp +#include +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 @@ -13,21 +24,167 @@ ## 答案 ```cpp - +class Solution +{ +public: + TreeNode *pre = NULL; + TreeNode *one = NULL; + TreeNode *two = NULL; + bool search(TreeNode *root) + { + if (root == NULL) + return false; + if (search(root->left)) + return true; + if (pre != NULL && pre->val > root->val) + { + if (one == NULL) + { + one = root; + two = pre; + } + else + { + two = root; + return true; + } + } + pre = root; + if (search(root->right)) + return true; + return false; + } + void recoverTree(TreeNode *root) + { + search(root); + swap(one->val, two->val); + } +}; ``` ## 选项 ### A ```cpp - +class Solution +{ +public: + TreeNode *x, *y, *pre; + void DFS(TreeNode *root) + { + if (root == nullptr) + return; + DFS(root->left); + if (pre && root->val < pre->val) + { + x = root; + if (!y) + y = pre; + else + return; + } + pre = root; + DFS(root->right); + } + void recoverTree(TreeNode *root) + { + DFS(root); + swap(x->val, y->val); + } +}; ``` ### B ```cpp - +class Solution +{ +public: + void insert(int val, vector &vals) + { + if (vals.size() > 0) + { + for (int i = 0; i < vals.size(); i++) + { + if (val < vals[i]) + { + vals.insert(vals.begin() + i, val); + return; + } + } + } + vals.push_back(val); + } + void recoverTree(TreeNode *root) + { + stack s; + vector vals; + TreeNode *tem = root; + while (tem || !s.empty()) + { + while (tem) + { + s.push(tem); + tem = tem->left; + } + if (!s.empty()) + { + tem = s.top(); + s.pop(); + insert(tem->val, vals); + tem = tem->right; + } + } + tem = root; + int j = 0; + while (tem || !s.empty()) + { + while (tem) + { + s.push(tem); + tem = tem->left; + } + if (!s.empty()) + { + tem = s.top(); + s.pop(); + tem->val = vals[j++]; + tem = tem->right; + } + } + } +}; ``` ### C ```cpp - +class Solution +{ +public: + vector order; + void inOrder(TreeNode *root) + { + if (root->left) + inOrder(root->left); + order.push_back(root); + if (root->right) + inOrder(root->right); + } + void recoverTree(TreeNode *root) + { + inOrder(root); + int left = -1, right = -1; + for (int i = 0; i < order.size() - 1; ++i) + { + if (order[i]->val > order[i + 1]->val) + { + if (left == -1) + left = i; + right = i + 1; + } + } + if (right == -1) + swap(order[left]->val, order[left + 1]->val); + else + swap(order[left]->val, order[right]->val); + } +}; ``` diff --git "a/data_backup/1.leetcode/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.md" "b/data_backup/1.leetcode/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.md" index 069816eb38c4cc91e5197e3acba70315234c5ab1..c58f0f3ecb4ff5179e5bf047974b64c14be21eb9 100644 --- "a/data_backup/1.leetcode/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.md" +++ "b/data_backup/1.leetcode/99_\347\233\270\345\220\214\347\232\204\346\240\221/solution.md" @@ -4,6 +4,18 @@ ## aop ### before ```cpp +#include +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 @@ -13,21 +25,66 @@ ## 答案 ```cpp - +都是错的 ``` ## 选项 ### A ```cpp - +class Solution +{ +public: + bool isSameTree(TreeNode *p, TreeNode *q) + { + if (p == nullptr) + return q == nullptr; + return q != nullptr && p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right); + } +}; ``` ### B ```cpp - +class Solution +{ +public: + bool isSameTree(TreeNode *p, TreeNode *q) + { + TreeNode *temp, *temq; + temp = p; + temq = q; + bool lb, rb; + if (!p && !q) + return true; + if ((!p || !q) || p->val != q->val) + return false; + else + { + lb = isSameTree(p->left, q->left); + rb = isSameTree(p->right, q->right); + } + if (lb && rb) + return true; + else + return false; + } +}; ``` ### C ```cpp - +class Solution +{ +public: + bool isSameTree(TreeNode *p, TreeNode *q) + { + if (p == NULL && q == NULL) + return true; + if (p == NULL || q == NULL) + return false; + if (p->val != q->val) + return false; + return (isSameTree(p->left, q->left) && isSameTree(p->right, q->right)); + } +}; ```