feat: add weekly contest 370 (doocs#1927)

* feat: add weekly contest 370
* chore: optimised images with calibre/image-actions

---------

Co-authored-by: Doocs Bot <doocs-bot@outlook.com>

Author: yanglbme

solution/0300-0399/0350.Intersection of Two Arrays II/README_EN.md

@@ -246,7 +246,6 @@ public class Solution {
 }
 ```
 
-
 ### **...**
 
 ```

solution/2900-2999/2923.Find Champion I/README.md

@@ -0,0 +1,92 @@
+# [2923. 找到冠军 I](https://leetcode.cn/problems/find-champion-i)
+
+[English Version](/solution/2900-2999/2923.Find%20Champion%20I/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>一场比赛中共有 <code>n</code> 支队伍，按从 <code>0</code> 到&nbsp; <code>n - 1</code> 编号。</p>
+
+<p>给你一个下标从 <strong>0</strong> 开始、大小为 <code>n * n</code> 的二维布尔矩阵 <code>grid</code> 。对于满足&nbsp;<code>0 &lt;= i, j &lt;= n - 1</code> 且 <code>i != j</code> 的所有 <code>i, j</code> ：如果 <code>grid[i][j] == 1</code>，那么 <code>i</code> 队比 <code>j</code> 队 <strong>强</strong> ；否则，<code>j</code> 队比 <code>i</code> 队 <strong>强</strong> 。</p>
+
+<p>在这场比赛中，如果不存在某支强于 <code>a</code> 队的队伍，则认为 <code>a</code> 队将会是 <strong>冠军</strong> 。</p>
+
+<p>返回这场比赛中将会成为冠军的队伍。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>grid = [[0,1],[0,0]]
+<strong>输出：</strong>0
+<strong>解释：</strong>比赛中有两支队伍。
+grid[0][1] == 1 表示 0 队比 1 队强。所以 0 队是冠军。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>grid = [[0,0,1],[1,0,1],[0,0,0]]
+<strong>输出：</strong>1
+<strong>解释：</strong>比赛中有三支队伍。
+grid[1][0] == 1 表示 1 队比 0 队强。
+grid[1][2] == 1 表示 1 队比 2 队强。
+所以 1 队是冠军。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>2 &lt;= n &lt;= 100</code></li>
+	<li><code>grid[i][j]</code> 的值为 <code>0</code> 或 <code>1</code></li>
+	<li>对于满足&nbsp;<code>i != j</code> 的所有 <code>i, j</code> ，<code>grid[i][j] != grid[j][i]</code> 均成立</li>
+	<li>生成的输出满足：如果 <code>a</code> 队比 <code>b</code> 队强，<code>b</code> 队比 <code>c</code> 队强，那么 <code>a</code> 队比 <code>c</code> 队强</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2923.Find Champion I/README_EN.md

@@ -0,0 +1,82 @@
+# [2923. Find Champion I](https://leetcode.com/problems/find-champion-i)
+
+[中文文档](/solution/2900-2999/2923.Find%20Champion%20I/README.md)
+
+## Description
+
+<p>There are <code>n</code> teams numbered from <code>0</code> to <code>n - 1</code> in a tournament.</p>
+
+<p>Given a <strong>0-indexed</strong> 2D boolean matrix <code>grid</code> of size <code>n * n</code>. For all <code>i, j</code> that <code>0 &lt;= i, j &lt;= n - 1</code> and <code>i != j</code> team <code>i</code> is <strong>stronger</strong> than team <code>j</code> if <code>grid[i][j] == 1</code>, otherwise, team <code>j</code> is <strong>stronger</strong> than team <code>i</code>.</p>
+
+<p>Team <code>a</code> will be the <strong>champion</strong> of the tournament if there is no team <code>b</code> that is stronger than team <code>a</code>.</p>
+
+<p>Return <em>the team that will be the champion of the tournament.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[0,1],[0,0]]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> There are two teams in this tournament.
+grid[0][1] == 1 means that team 0 is stronger than team 1. So team 0 will be the champion.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[0,0,1],[1,0,1],[0,0,0]]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> There are three teams in this tournament.
+grid[1][0] == 1 means that team 1 is stronger than team 0.
+grid[1][2] == 1 means that team 1 is stronger than team 2.
+So team 1 will be the champion.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>2 &lt;= n &lt;= 100</code></li>
+	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
+	<li>For all <code>i, j</code> that <code>i != j</code>, <code>grid[i][j] != grid[j][i]</code>.</li>
+	<li>The input is generated such that if team <code>a</code> is stronger than team <code>b</code> and team <code>b</code> is stronger than team <code>c</code>, then team <code>a</code> is stronger than team <code>c</code>.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2924.Find Champion II/README.md

@@ -0,0 +1,103 @@
+# [2924. 找到冠军 II](https://leetcode.cn/problems/find-champion-ii)
+
+[English Version](/solution/2900-2999/2924.Find%20Champion%20II/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>一场比赛中共有 <code>n</code> 支队伍，按从 <code>0</code> 到&nbsp; <code>n - 1</code> 编号。每支队伍也是 <strong>有向无环图（DAG）</strong> 上的一个节点。</p>
+
+<p>给你一个整数 <code>n</code> 和一个下标从 <strong>0</strong> 开始、长度为 <code>m</code> 的二维整数数组 <code>edges</code> 表示这个有向无环图，其中 <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> 表示图中存在一条从 <code>u<sub>i</sub></code> 队到 <code>v<sub>i</sub></code> 队的有向边。</p>
+
+<p>从 <code>a</code> 队到 <code>b</code> 队的有向边意味着 <code>a</code> 队比 <code>b</code> 队 <strong>强</strong> ，也就是 <code>b</code> 队比 <code>a</code> 队 <strong>弱</strong> 。</p>
+
+<p>在这场比赛中，如果不存在某支强于 <code>a</code> 队的队伍，则认为 <code>a</code> 队将会是 <strong>冠军</strong> 。</p>
+
+<p>如果这场比赛存在 <strong>唯一</strong> 一个冠军，则返回将会成为冠军的队伍。否则，返回<em> </em><code>-1</code><em> 。</em></p>
+
+<p><strong>注意</strong></p>
+
+<ul>
+	<li><strong>环</strong> 是形如 <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub>, a<sub>n+1</sub></code> 的一个序列，且满足：节点 <code>a<sub>1</sub></code> 与节点 <code>a<sub>n+1</sub></code> 是同一个节点；节点 <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub></code> 互不相同；对于范围&nbsp;<code>[1, n]</code> 中的每个 <code>i</code> ，均存在一条从节点 <code>a<sub>i</sub></code> 到节点 <code>a<sub>i+1</sub></code> 的有向边。</li>
+	<li><strong>有向无环图</strong> 是不存在任何环的有向图。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<p><img height="300" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2924.Find%20Champion%20II/images/graph-3.png" width="300" /></p>
+
+<pre>
+<strong>输入：</strong>n = 3, edges = [[0,1],[1,2]]
+<strong>输出：</strong>0
+<strong>解释：</strong>1 队比 0 队弱。2 队比 1 队弱。所以冠军是 0 队。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<p><img height="300" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2924.Find%20Champion%20II/images/graph-4.png" width="300" /></p>
+
+<pre>
+<strong>输入：</strong>n = 4, edges = [[0,2],[1,3],[1,2]]
+<strong>输出：</strong>-1
+<strong>解释：</strong>2 队比 0 队和 1 队弱。3 队比 1 队弱。但是 1 队和 0 队之间不存在强弱对比。所以答案是 -1 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>m == edges.length</code></li>
+	<li><code>0 &lt;= m &lt;= n * (n - 1) / 2</code></li>
+	<li><code>edges[i].length == 2</code></li>
+	<li><code>0 &lt;= edge[i][j] &lt;= n - 1</code></li>
+	<li><code>edges[i][0] != edges[i][1]</code></li>
+	<li>生成的输入满足：如果 <code>a</code> 队比 <code>b</code> 队强，就不存在 <code>b</code> 队比 <code>a</code> 队强</li>
+	<li>生成的输入满足：如果 <code>a</code> 队比 <code>b</code> 队强，<code>b</code> 队比 <code>c</code> 队强，那么 <code>a</code> 队比 <code>c</code> 队强</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2924.Find Champion II/README_EN.md

@@ -0,0 +1,93 @@
+# [2924. Find Champion II](https://leetcode.com/problems/find-champion-ii)
+
+[中文文档](/solution/2900-2999/2924.Find%20Champion%20II/README.md)
+
+## Description
+
+<p>There are <code>n</code> teams numbered from <code>0</code> to <code>n - 1</code> in a tournament; each team is also a node in a <strong>DAG</strong>.</p>
+
+<p>You are given the integer <code>n</code> and a <strong>0-indexed</strong> 2D integer array <code>edges</code> of length <code><font face="monospace">m</font></code> representing the <strong>DAG</strong>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is a directed edge from team <code>u<sub>i</sub></code> to team <code>v<sub>i</sub></code> in the graph.</p>
+
+<p>A directed edge from <code>a</code> to <code>b</code> in the graph means that team <code>a</code> is <strong>stronger</strong> than team <code>b</code> and team <code>b</code> is <strong>weaker</strong> than team <code>a</code>.</p>
+
+<p>Team <code>a</code> will be the <strong>champion</strong> of the tournament if there is no team <code>b</code> that is <strong>stronger</strong> than team <code>a</code>.</p>
+
+<p>Return <em>the team that will be the <strong>champion</strong> of the tournament if there is a <strong>unique</strong> champion, otherwise, return </em><code>-1</code><em>.</em></p>
+
+<p><strong>Notes</strong></p>
+
+<ul>
+	<li>A <strong>cycle</strong> is a series of nodes <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub>, a<sub>n+1</sub></code> such that node <code>a<sub>1</sub></code> is the same node as node <code>a<sub>n+1</sub></code>, the nodes <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub></code> are distinct, and there is a directed edge from the node <code>a<sub>i</sub></code> to node <code>a<sub>i+1</sub></code> for every <code>i</code> in the range <code>[1, n]</code>.</li>
+	<li>A <strong>DAG</strong> is a directed graph that does not have any <strong>cycle</strong>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<p><img height="300" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2924.Find%20Champion%20II/images/graph-3.png" width="300" /></p>
+
+<pre>
+<strong>Input:</strong> n = 3, edges = [[0,1],[1,2]]
+<strong>Output:</strong> 0
+<strong>Explanation: </strong>Team 1 is weaker than team 0. Team 2 is weaker than team 1. So the champion is team 0.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<p><img height="300" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2924.Find%20Champion%20II/images/graph-4.png" width="300" /></p>
+
+<pre>
+<strong>Input:</strong> n = 4, edges = [[0,2],[1,3],[1,2]]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> Team 2 is weaker than team 0 and team 1. Team 3 is weaker than team 1. But team 1 and team 0 are not weaker than any other teams. So the answer is -1.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>m == edges.length</code></li>
+	<li><code>0 &lt;= m &lt;= n * (n - 1) / 2</code></li>
+	<li><code>edges[i].length == 2</code></li>
+	<li><code>0 &lt;= edge[i][j] &lt;= n - 1</code></li>
+	<li><code>edges[i][0] != edges[i][1]</code></li>
+	<li>The input is generated such that if team <code>a</code> is stronger than team <code>b</code>, team <code>b</code> is not stronger than team <code>a</code>.</li>
+	<li>The input is generated such that if team <code>a</code> is stronger than team <code>b</code> and team <code>b</code> is stronger than team <code>c</code>, then team <code>a</code> is stronger than team <code>c</code>.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->