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data/1.dailycode初阶/2.java/101.exercises/solution.json → data/1.dailycode初阶/2.java/49.exercises/solution.json
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data/1.dailycode初阶/2.java/104.exercises/solution.json → data/1.dailycode初阶/2.java/50.exercises/solution.json
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data/1.dailycode初阶/2.java/108.exercises/solution.json → data/1.dailycode初阶/2.java/51.exercises/solution.json
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data/1.dailycode初阶/2.java/110.exercises/solution.json → data/1.dailycode初阶/2.java/52.exercises/solution.json
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data/1.dailycode初阶/2.java/111.exercises/solution.json → data/1.dailycode初阶/2.java/53.exercises/solution.json
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data/1.dailycode初阶/2.java/112.exercises/solution.json → data/1.dailycode初阶/2.java/54.exercises/solution.json
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data/1.dailycode初阶/2.java/118.exercises/solution.json → data/1.dailycode初阶/2.java/55.exercises/solution.json
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data/1.dailycode初阶/2.java/119.exercises/solution.json → data/1.dailycode初阶/2.java/56.exercises/solution.json
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data/1.dailycode初阶/2.java/121.exercises/solution.json → data/1.dailycode初阶/2.java/57.exercises/solution.json
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data/1.dailycode初阶/2.java/122.exercises/solution.json → data/1.dailycode初阶/2.java/58.exercises/solution.json
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data/1.dailycode初阶/2.java/125.exercises/solution.json → data/1.dailycode初阶/2.java/59.exercises/solution.json
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data/1.dailycode初阶/2.java/136.exercises/solution.json → data/1.dailycode初阶/2.java/60.exercises/solution.json
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......@@ -3,5 +3,6 @@
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data/1.dailycode初阶/2.java/141.exercises/solution.json → data/1.dailycode初阶/2.java/61.exercises/solution.json
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data/1.dailycode初阶/2.java/144.exercises/solution.json → data/1.dailycode初阶/2.java/62.exercises/solution.json
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data/1.dailycode初阶/2.java/145.exercises/solution.json → data/1.dailycode初阶/2.java/63.exercises/solution.json
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data/1.dailycode初阶/2.java/155.exercises/solution.json → data/1.dailycode初阶/2.java/64.exercises/solution.json
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data/1.dailycode初阶/2.java/167.exercises/solution.json → data/1.dailycode初阶/2.java/65.exercises/solution.json
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data/1.dailycode初阶/2.java/168.exercises/solution.json → data/1.dailycode初阶/2.java/66.exercises/solution.json
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data/1.dailycode初阶/2.java/171.exercises/solution.json → data/1.dailycode初阶/2.java/68.exercises/solution.json
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data/1.dailycode初阶/2.java/172.exercises/solution.json → data/1.dailycode初阶/2.java/69.exercises/solution.json
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data/1.dailycode初阶/2.java/191.exercises/solution.json → data/1.dailycode初阶/2.java/71.exercises/solution.json
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data/1.dailycode初阶/2.java/202.exercises/solution.json → data/1.dailycode初阶/2.java/72.exercises/solution.json
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data/1.dailycode初阶/2.java/203.exercises/solution.json → data/1.dailycode初阶/2.java/73.exercises/solution.json
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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data/1.dailycode初阶/3.python/104.exercises/solution.json → data/1.dailycode初阶/3.python/50.exercises/solution.json
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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data/1.dailycode初阶/3.python/155.exercises/solution.json → data/1.dailycode初阶/3.python/64.exercises/solution.json
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
"source": "solution.md",
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"source": "solution.md",
"exercise_id": "6b2211be24844846afddd9cec234f4e5",
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"keywords": "数学"
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......@@ -3,5 +3,6 @@
"source": "solution.md",
"exercise_id": "1ba368a03960439599d6d1b5e6867284",
"author": "csdn.net",
"keywords": "位运算,分治"
"keywords": "位运算,分治",
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
"source": "solution.md",
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......@@ -3,5 +3,6 @@
"source": "solution.md",
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
"source": "solution.md",
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......@@ -3,5 +3,6 @@
"source": "solution.md",
"exercise_id": "1846ac34e8624af7b4080f3b6af1b5aa",
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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data/2.dailycode中阶/2.java/102.exercises/config.json
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......
data/2.dailycode中阶/2.java/102.exercises/solution.json
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{
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data/2.dailycode中阶/2.java/102.exercises/solution.md
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# 二叉树的层序遍历
# 二叉树的右视图
<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>给定一个二叉树的 <strong>根节点</strong> <code>root</code>，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。</p>
<p> </p>
<p><strong>示例：</strong><br />
二叉树：<code>[3,9,20,null,null,15,7]</code>,</p>
<p><strong>示例 1:</strong></p>
<p><img src="https://assets.leetcode.com/uploads/2021/02/14/tree.jpg" style="width: 270px; " /></p>
<pre>
3
/ \
9 20
/ \
15 7
<strong>输入:</strong> [1,2,3,null,5,null,4]
<strong>输出:</strong> [1,3,4]
</pre>
<p>返回其层序遍历结果：</p>
<p><strong>示例 2:</strong></p>
<pre>
[
[3],
[9,20],
[15,7]
]
<strong>输入:</strong> [1,null,3]
<strong>输出:</strong> [1,3]
</pre>
<p><strong>示例 3:</strong></p>
<pre>
<strong>输入:</strong> []
<strong>输出:</strong> []
</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li>二叉树的节点个数的范围是 <code>[0,100]</code></li>
<li><meta charset="UTF-8" /><code>-100 <= Node.val <= 100</code> </li>
</ul>
## template
......@@ -40,29 +51,24 @@ public class TreeNode {
}
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> l = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<TreeNode>();
if (root != null) {
q.add(root);
}
while (!q.isEmpty()) {
List<Integer> l2 = new ArrayList<>();
int number = q.size();
while (number > 0) {
TreeNode t = q.poll();
l2.add(t.val);
if (t.left != null) {
q.add(t.left);
}
if (t.right != null) {
q.add(t.right);
}
number--;
public List<Integer> rightSideView(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if (root == null)
return list;
list.add(root.val);
List<Integer> list1 = rightSideView(root.right);
List<Integer> list2 = rightSideView(root.left);
list.addAll(list1);
if (list1.size() < list2.size()) {
for (int i = list1.size(); i < list2.size(); i++) {
list.add(list2.get(i));
}
l.add(l2);
}
return l;
return list;
}
}
......
data/2.dailycode中阶/2.java/103.exercises/config.json
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{
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"node_id": "dailycode-805cf116376a4e7bbb4b28835ecb711d",
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......
data/2.dailycode中阶/2.java/103.exercises/solution.json
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"notebook_enable": true
}
\ No newline at end of file
data/2.dailycode中阶/2.java/103.exercises/solution.md
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# 二叉树的锯齿形层序遍历
# 岛屿数量
<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>给你一个由 <code>'1'</code>（陆地）和 <code>'0'</code>（水）组成的的二维网格，请你计算网格中岛屿的数量。</p>
<p>例如：<br />
给定二叉树 <code>[3,9,20,null,null,15,7]</code>,</p>
<p>岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。</p>
<p>此外，你可以假设该网格的四条边均被水包围。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
3
/ \
9 20
/ \
15 7
<strong>输入：</strong>grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
<strong>输出：</strong>1
</pre>
<p>返回锯齿形层序遍历如下：</p>
<p><strong>示例 2：</strong></p>
<pre>
[
[3],
[20,9],
[15,7]
<strong>输入：</strong>grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
<strong>输出：</strong>3
</pre>
<p> </p>
## template
<p><strong>提示：</strong></p>
```java
<ul>
<li><code>m == grid.length</code></li>
<li><code>n == grid[i].length</code></li>
<li><code>1 <= m, n <= 300</code></li>
<li><code>grid[i][j]</code> 的值为 <code>'0'</code> 或 <code>'1'</code></li>
</ul>
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
## template
```java
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> list = new LinkedList<>();
if (root == null) {
return list;
}
Stack<TreeNode> stack1 = new Stack<>();
stack1.push(root);
boolean postive = true;
while (!stack1.isEmpty()) {
Stack<TreeNode> stack2 = new Stack<>();
List<Integer> subList = new LinkedList<>();
while (!stack1.isEmpty()) {
TreeNode current = stack1.pop();
subList.add(current.val);
if (postive) {
if (current.left != null) {
stack2.push(current.left);
}
if (current.right != null) {
stack2.push(current.right);
}
} else {
if (current.right != null) {
stack2.push(current.right);
}
if (current.left != null) {
stack2.push(current.left);
}
public int numIslands(char[][] grid) {
int islandNum = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '1') {
infect(grid, i, j);
islandNum++;
}
}
postive = !postive;
stack1 = stack2;
list.add(subList);
}
return islandNum;
}
return list;
public void infect(char[][] grid, int i, int j) {
if (i < 0 || i >= grid.length ||
j < 0 || j >= grid[0].length || grid[i][j] != '1') {
return;
}
grid[i][j] = '2';
infect(grid, i + 1, j);
infect(grid, i - 1, j);
infect(grid, i, j + 1);
infect(grid, i, j - 1);
}
}
```
## 答案
......
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{
"node_id": "dailycode-aafb162733204b54ad1659371e1b8f4a",
"keywords": [],
"children": [],
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"solution.json"
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"source": "solution.md",
"exercise_id": "b68c4b0a4f974ab4b38ec8d3a8eb9c77",
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"keywords": "树,深度优先搜索,广度优先搜索,二叉树"
}
\ No newline at end of file
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# 二叉树的右视图
<p>给定一个二叉树的 <strong>根节点</strong> <code>root</code>，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。</p>
<p> </p>
<p><strong>示例 1:</strong></p>
<p><img src="https://assets.leetcode.com/uploads/2021/02/14/tree.jpg" style="width: 270px; " /></p>
<pre>
<strong>输入:</strong> [1,2,3,null,5,null,4]
<strong>输出:</strong> [1,3,4]
</pre>
<p><strong>示例 2:</strong></p>
<pre>
<strong>输入:</strong> [1,null,3]
<strong>输出:</strong> [1,3]
</pre>
<p><strong>示例 3:</strong></p>
<pre>
<strong>输入:</strong> []
<strong>输出:</strong> []
</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li>二叉树的节点个数的范围是 <code>[0,100]</code></li>
<li><meta charset="UTF-8" /><code>-100 <= Node.val <= 100</code> </li>
</ul>
## template
```java
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if (root == null)
return list;
list.add(root.val);
List<Integer> list1 = rightSideView(root.right);
List<Integer> list2 = rightSideView(root.left);
list.addAll(list1);
if (list1.size() < list2.size()) {
for (int i = list1.size(); i < list2.size(); i++) {
list.add(list2.get(i));
}
}
return list;
}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```
\ No newline at end of file
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{
"node_id": "dailycode-805cf116376a4e7bbb4b28835ecb711d",
"keywords": [],
"children": [],
"keywords_must": [],
"keywords_forbid": [],
"export": [
"solution.json"
]
}
\ No newline at end of file
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{
"type": "code_options",
"source": "solution.md",
"exercise_id": "e7be0db8bb5844eba148a6e28f727ae3",
"author": "csdn.net",
"keywords": "深度优先搜索,广度优先搜索,并查集,数组,矩阵"
}
\ No newline at end of file
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# 岛屿数量
<p>给你一个由 <code>'1'</code>（陆地）和 <code>'0'</code>（水）组成的的二维网格，请你计算网格中岛屿的数量。</p>
<p>岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。</p>
<p>此外，你可以假设该网格的四条边均被水包围。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
<strong>输出：</strong>1
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
<strong>输出：</strong>3
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>m == grid.length</code></li>
<li><code>n == grid[i].length</code></li>
<li><code>1 <= m, n <= 300</code></li>
<li><code>grid[i][j]</code> 的值为 <code>'0'</code> 或 <code>'1'</code></li>
</ul>
## template
```java
class Solution {
public int numIslands(char[][] grid) {
int islandNum = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '1') {
infect(grid, i, j);
islandNum++;
}
}
}
return islandNum;
}
public void infect(char[][] grid, int i, int j) {
if (i < 0 || i >= grid.length ||
j < 0 || j >= grid[0].length || grid[i][j] != '1') {
return;
}
grid[i][j] = '2';
infect(grid, i + 1, j);
infect(grid, i - 1, j);
infect(grid, i, j + 1);
infect(grid, i, j - 1);
}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```
\ No newline at end of file
data/2.dailycode中阶/2.java/65.exercises/config.json 0 → 100644
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{
"node_id": "dailycode-a2ccc65d66f7450e8405eefda34c259d",
"keywords": [],
"children": [],
"keywords_must": [],
"keywords_forbid": [],
"export": [
"solution.json"
]
}
\ No newline at end of file
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{
"type": "code_options",
"source": "solution.md",
"exercise_id": "682f41a955f04da9bb0feb3d063f8b3e",
"author": "csdn.net",
"keywords": "树,广度优先搜索,二叉树",
"notebook_enable": true
}
\ No newline at end of file
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# 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>
## template
```java
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> l = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<TreeNode>();
if (root != null) {
q.add(root);
}
while (!q.isEmpty()) {
List<Integer> l2 = new ArrayList<>();
int number = q.size();
while (number > 0) {
TreeNode t = q.poll();
l2.add(t.val);
if (t.left != null) {
q.add(t.left);
}
if (t.right != null) {
q.add(t.right);
}
number--;
}
l.add(l2);
}
return l;
}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```
\ No newline at end of file
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{
"node_id": "dailycode-a10ff4a447fb49fca996a9c4c95b4991",
"keywords": [],
"children": [],
"keywords_must": [],
"keywords_forbid": [],
"export": [
"solution.json"
]
}
\ No newline at end of file
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{
"type": "code_options",
"source": "solution.md",
"exercise_id": "622d6183d44744c68b49f96bb9c87452",
"author": "csdn.net",
"keywords": "树,广度优先搜索,二叉树",
"notebook_enable": true
}
\ No newline at end of file
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# 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>
## template
```java
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> list = new LinkedList<>();
if (root == null) {
return list;
}
Stack<TreeNode> stack1 = new Stack<>();
stack1.push(root);
boolean postive = true;
while (!stack1.isEmpty()) {
Stack<TreeNode> stack2 = new Stack<>();
List<Integer> subList = new LinkedList<>();
while (!stack1.isEmpty()) {
TreeNode current = stack1.pop();
subList.add(current.val);
if (postive) {
if (current.left != null) {
stack2.push(current.left);
}
if (current.right != null) {
stack2.push(current.right);
}
} else {
if (current.right != null) {
stack2.push(current.right);
}
if (current.left != null) {
stack2.push(current.left);
}
}
}
postive = !postive;
stack1 = stack2;
list.add(subList);
}
return list;
}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```
\ No newline at end of file
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......@@ -3,5 +3,6 @@
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"source": "solution.md",
"exercise_id": "12c18413a51d4dc0aaff19498b606d12",
"author": "csdn.net",
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"notebook_enable": true
}
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......@@ -3,5 +3,6 @@
"source": "solution.md",
"exercise_id": "c243d27f11ae40178101f3a631183c6e",
"author": "csdn.net",
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"keywords": "树,广度优先搜索,二叉树",
"notebook_enable": true
}
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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......@@ -3,5 +3,6 @@
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"exercise_id": "6f1eb2d1c9aa4dc6a97e7c8076573729",
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......@@ -3,5 +3,6 @@
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"notebook_enable": true
}
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......@@ -3,5 +3,6 @@
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"exercise_id": "636abfea1a2b4607859af3c78cb96e84",
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}
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data/2.dailycode中阶/2.java/128.exercises/solution.json → data/2.dailycode中阶/2.java/76.exercises/solution.json
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......@@ -3,5 +3,6 @@
"source": "solution.md",
"exercise_id": "17afc370460f46c0adbaab615712518f",
"author": "csdn.net",
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"notebook_enable": true
}
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......@@ -3,5 +3,6 @@
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"exercise_id": "4ae056aaec24487aa6c7c5664dcd9106",
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"keywords": "树,深度优先搜索,二叉树",
"notebook_enable": true
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......@@ -3,5 +3,6 @@
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"exercise_id": "be34956c33dd4d54bbed87721b778ad6",
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"exercise_id": "d946d51d15644bc6b5147e846d0a12d4",
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"author": "csdn.net",
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"source": "solution.md",
"exercise_id": "0e368bf69c094190966ba3515e36d864",
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"source": "solution.md",
"exercise_id": "e122bca4653a4a05a64157c9d3afb931",
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"keywords": "哈希表,链表,双指针",
"notebook_enable": true
}
\ No newline at end of file
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"source": "solution.md",
"exercise_id": "197a8b12e097460ab3a266818f16699c",
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"source": "solution.md",
"exercise_id": "fcd911bde6db4e1f8ba109f6e3dd6a7e",
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"exercise_id": "6267d9fe114f48b890c6ed28dc63da05",
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"keywords": "链表,排序"
"keywords": "链表,排序",
"notebook_enable": true
}
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"source": "solution.md",
"exercise_id": "71419e1d7bc74b23958e1b78923ad9a1",
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"notebook_enable": true
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"exercise_id": "30c3029500da406d97c9f663305f3a22",
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"keywords": "栈,数组,数学",
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"source": "solution.md",
"exercise_id": "2a8b52e0c9894ceea4d669185da8bafa",
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"source": "solution.md",
"exercise_id": "65950a04beae457f82920300e5028918",
"author": "csdn.net",
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"keywords": "数组,动态规划",
"notebook_enable": true
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"source": "solution.md",
"exercise_id": "27c7eb432aa541659f91e42739ce47c5",
"author": "csdn.net",
"keywords": "数组,二分查找"
"keywords": "数组,二分查找",
"notebook_enable": true
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"source": "solution.md",
"exercise_id": "777e3a0d404c492096839996e9bd0388",
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"keywords": "数组,二分查找",
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"source": "solution.md",
"exercise_id": "244c65fcb46345da8a6f79f4a63934ef",
"author": "csdn.net",
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}
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"source": "solution.md",
"exercise_id": "1ff02b73fbc84fdca3e87f23ac478ce1",
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"keywords": "哈希表,数学,字符串"
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"notebook_enable": true
}
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"source": "solution.md",
"exercise_id": "e18c19c6ef0d4268a8154c9b3e1a2f1a",
"author": "csdn.net",
"keywords": "栈,树,设计,二叉搜索树,二叉树,迭代器"
"keywords": "栈,树,设计,二叉搜索树,二叉树,迭代器",
"notebook_enable": true
}
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"source": "solution.md",
"exercise_id": "694715c7a1484e80b939cbaed3256af1",
"author": "csdn.net",
"keywords": "贪心,字符串,排序"
"keywords": "贪心,字符串,排序",
"notebook_enable": true
}
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"source": "solution.md",
"exercise_id": "84d39ae3a62d48309d5c5dda130a901f",
"author": "csdn.net",
"keywords": "位运算,哈希表,字符串,滑动窗口,哈希函数,滚动哈希"
"keywords": "位运算,哈希表,字符串,滑动窗口,哈希函数,滚动哈希",
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}
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"source": "solution.md",
"exercise_id": "ea8fa229d3c243a4a6f6ebfdccdd5bfd",
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"keywords": "数组,动态规划",
"notebook_enable": true
}
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"source": "solution.md",
"exercise_id": "b1ba5727356c443b859bfb64fc002898",
"author": "csdn.net",
"keywords": "树,深度优先搜索,广度优先搜索,二叉树"
"keywords": "树,深度优先搜索,广度优先搜索,二叉树",
"notebook_enable": true
}
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data/2.dailycode中阶/3.python/102.exercises/config.json
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{
"node_id": "dailycode-d39c19f3422842d4bd0ec6adf5a12e78",
"node_id": "dailycode-72651cf7970c4a9c837aee8016c169d1",
"keywords": [],
"children": [],
"keywords_must": [],
......
data/2.dailycode中阶/3.python/102.exercises/solution.json
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{
"type": "code_options",
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"exercise_id": "6e676186276f4879bfcd9fb546041c14",
"exercise_id": "22a947b155ff40e9825fdf1bb533aa04",
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"keywords": "深度优先搜索,广度优先搜索,并查集,数组,矩阵",
"notebook_enable": true
}
\ No newline at end of file
data/2.dailycode中阶/3.python/102.exercises/solution.md
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# 二叉树的层序遍历
# 岛屿数量
<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>给你一个由 <code>'1'</code>（陆地）和 <code>'0'</code>（水）组成的的二维网格，请你计算网格中岛屿的数量。</p>
<p>岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。</p>
<p>此外，你可以假设该网格的四条边均被水包围。</p>
<p> </p>
<p><strong>示例：</strong><br />
二叉树：<code>[3,9,20,null,null,15,7]</code>,</p>
<p><strong>示例 1：</strong></p>
<pre>
3
/ \
9 20
/ \
15 7
<strong>输入：</strong>grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
<strong>输出：</strong>1
</pre>
<p>返回其层序遍历结果：</p>
<p><strong>示例 2：</strong></p>
<pre>
[
[3],
[9,20],
[15,7]
<strong>输入：</strong>grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
<strong>输出：</strong>3
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>m == grid.length</code></li>
<li><code>n == grid[i].length</code></li>
<li><code>1 <= m, n <= 300</code></li>
<li><code>grid[i][j]</code> 的值为 <code>'0'</code> 或 <code>'1'</code></li>
</ul>
## template
```python
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
queue, res = [root], []
while queue:
size = len(queue)
temp = []
for i in range(size):
data = queue.pop(0)
temp.append(data.val)
if data.left:
queue.append(data.left)
if data.right:
queue.append(data.right)
res.append(temp)
return res
class Solution:
def depthSearch(self, grid, i, j):
if i < 0 or j < 0 or i >= len(grid) or j >= len(grid[0]):
return
if grid[i][j] == "0":
return
grid[i][j] = "0"
self.depthSearch(grid, i - 1, j)
self.depthSearch(grid, i + 1, j)
self.depthSearch(grid, i, j - 1)
self.depthSearch(grid, i, j + 1)
def numIslands(self, grid: List[List[str]]) -> int:
if not grid:
return 0
w, h = len(grid[0]), len(grid)
cnt = 0
for i in range(h):
for j in range(w):
if grid[i][j] == "1":
self.depthSearch(grid, i, j)
cnt += 1
return cnt
```
......
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{
"node_id": "dailycode-72651cf7970c4a9c837aee8016c169d1",
"keywords": [],
"children": [],
"keywords_must": [],
"keywords_forbid": [],
"export": [
"solution.json"
]
}
\ No newline at end of file
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{
"type": "code_options",
"source": "solution.md",
"exercise_id": "22a947b155ff40e9825fdf1bb533aa04",
"author": "csdn.net",
"keywords": "深度优先搜索,广度优先搜索,并查集,数组,矩阵"
}
\ No newline at end of file
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# 岛屿数量
<p>给你一个由 <code>'1'</code>（陆地）和 <code>'0'</code>（水）组成的的二维网格，请你计算网格中岛屿的数量。</p>
<p>岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。</p>
<p>此外，你可以假设该网格的四条边均被水包围。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
<strong>输出：</strong>1
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
<strong>输出：</strong>3
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>m == grid.length</code></li>
<li><code>n == grid[i].length</code></li>
<li><code>1 <= m, n <= 300</code></li>
<li><code>grid[i][j]</code> 的值为 <code>'0'</code> 或 <code>'1'</code></li>
</ul>
## template
```python
class Solution:
def depthSearch(self, grid, i, j):
if i < 0 or j < 0 or i >= len(grid) or j >= len(grid[0]):
return
if grid[i][j] == "0":
return
grid[i][j] = "0"
self.depthSearch(grid, i - 1, j)
self.depthSearch(grid, i + 1, j)
self.depthSearch(grid, i, j - 1)
self.depthSearch(grid, i, j + 1)
def numIslands(self, grid: List[List[str]]) -> int:
if not grid:
return 0
w, h = len(grid[0]), len(grid)
cnt = 0
for i in range(h):
for j in range(w):
if grid[i][j] == "1":
self.depthSearch(grid, i, j)
cnt += 1
return cnt
```
## 答案
```python
```
## 选项
### A
```python
```
### B
```python
```
### C
```python
```
\ No newline at end of file
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{
"node_id": "dailycode-d39c19f3422842d4bd0ec6adf5a12e78",
"keywords": [],
"children": [],
"keywords_must": [],
"keywords_forbid": [],
"export": [
"solution.json"
]
}
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{
"type": "code_options",
"source": "solution.md",
"exercise_id": "6e676186276f4879bfcd9fb546041c14",
"author": "csdn.net",
"keywords": "树,广度优先搜索,二叉树",
"notebook_enable": true
}
\ No newline at end of file
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# 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>
## template
```python
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
queue, res = [root], []
while queue:
size = len(queue)
temp = []
for i in range(size):
data = queue.pop(0)
temp.append(data.val)
if data.left:
queue.append(data.left)
if data.right:
queue.append(data.right)
res.append(temp)
return res
```
## 答案
```python
```
## 选项
### A
```python
```
### B
```python
```
### C
```python
```
\ No newline at end of file
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data/tree.json
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因为 它太大了无法显示 source diff 。你可以改为 查看blob。
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