feat: add weekly contest 379 (#2191)

Author: acbin

solution/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README.md

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+# [10035. 对角线最长的矩形的面积](https://leetcode.cn/problems/maximum-area-of-longest-diagonal-rectangle)
+
+[English Version](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个下标从<strong> 0</strong> 开始的二维整数数组 <code>dimensions</code>。</p>
+
+<p>对于所有下标 <code>i</code>（<code>0 &lt;= i &lt; dimensions.length</code>），<code>dimensions[i][0]</code> 表示矩形 <span style="font-size: 13.3333px;"> <code>i</code></span> 的长度，而 <code>dimensions[i][1]</code> 表示矩形 <span style="font-size: 13.3333px;"> <code>i</code></span> 的宽度。</p>
+
+<p>返回对角线最 <strong>长 </strong>的矩形的<strong> 面积 </strong>。如果存在多个对角线长度相同的矩形，返回面积最<strong> 大 </strong>的矩形的面积。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>dimensions = [[9,3],[8,6]]
+<strong>输出：</strong>48
+<strong>解释：</strong>
+下标 = 0，长度 = 9，宽度 = 3。对角线长度 = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈<!-- notionvc: 882cf44c-3b17-428e-9c65-9940810216f1 --> 9.487。
+下标 = 1，长度 = 8，宽度 = 6。对角线长度 = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10。
+因此，下标为 1 的矩形对角线更长，所以返回面积 = 8 * 6 = 48。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>dimensions = [[3,4],[4,3]]
+<strong>输出：</strong>12
+<strong>解释：</strong>两个矩形的对角线长度相同，为 5，所以最大面积 = 12。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= dimensions.length &lt;= 100</code></li>
+	<li><code>dimensions[i].length == 2</code></li>
+	<li><code>1 &lt;= dimensions[i][0], dimensions[i][1] &lt;= 100</code></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/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README_EN.md

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+# [10035. Maximum Area of Longest Diagonal Rectangle](https://leetcode.com/problems/maximum-area-of-longest-diagonal-rectangle)
+
+[中文文档](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README.md)
+
+## Description
+
+<p>You are given a 2D <strong>0-indexed </strong>integer array <code>dimensions</code>.</p>
+
+<p>For all indices <code>i</code>, <code>0 &lt;= i &lt; dimensions.length</code>, <code>dimensions[i][0]</code> represents the length and <code>dimensions[i][1]</code> represents the width of the rectangle<span style="font-size: 13.3333px;"> <code>i</code></span>.</p>
+
+<p>Return <em>the <strong>area</strong> of the rectangle having the <strong>longest</strong> diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the <strong>maximum</strong> area.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> dimensions = [[9,3],[8,6]]
+<strong>Output:</strong> 48
+<strong>Explanation:</strong> 
+For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) &asymp;<!-- notionvc: 882cf44c-3b17-428e-9c65-9940810216f1 --> 9.487.
+For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
+So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> dimensions = [[3,4],[4,3]]
+<strong>Output:</strong> 12
+<strong>Explanation:</strong> Length of diagonal is the same for both which is 5, so maximum area = 12.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= dimensions.length &lt;= 100</code></li>
+	<li><code><font face="monospace">dimensions[i].length == 2</font></code></li>
+	<li><code><font face="monospace">1 &lt;= dimensions[i][0], dimensions[i][1] &lt;= 100</font></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/10000-10099/10036.Minimum Moves to Capture The Queen/README.md

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+# [10036. 捕获黑皇后需要的最少移动次数](https://leetcode.cn/problems/minimum-moves-to-capture-the-queen)
+
+[English Version](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>现有一个下标从 <strong>0</strong> 开始的 <code>8 x 8</code> 棋盘，上面有 <code>3</code> 枚棋子。</p>
+
+<p>给你 <code>6</code> 个整数 <code>a</code> 、<code>b</code> 、<code>c</code> 、<code>d</code> 、<code>e</code> 和 <code>f</code> ，其中：</p>
+
+<ul>
+	<li><code>(a, b)</code> 表示白色车的位置。</li>
+	<li><code>(c, d)</code> 表示白色象的位置。</li>
+	<li><code>(e, f)</code> 表示黑皇后的位置。</li>
+</ul>
+
+<p>假定你只能移动白色棋子，返回捕获黑皇后所需的<strong>最少</strong>移动次数。</p>
+
+<p><strong>请注意</strong>：</p>
+
+<ul>
+	<li>车可以向垂直或水平方向移动任意数量的格子，但不能跳过其他棋子。</li>
+	<li>象可以沿对角线方向移动任意数量的格子，但不能跳过其他棋子。</li>
+	<li>如果车或象能移向皇后所在的格子，则认为它们可以捕获皇后。</li>
+	<li>皇后不能移动。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/images//ex1.png" style="width: 600px; height: 600px; padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>输入：</strong>a = 1, b = 1, c = 8, d = 8, e = 2, f = 3
+<strong>输出：</strong>2
+<strong>解释：</strong>将白色车先移动到 (1, 3) ，然后移动到 (2, 3) 来捕获黑皇后，共需移动 2 次。
+由于起始时没有任何棋子正在攻击黑皇后，要想捕获黑皇后，移动次数不可能少于 2 次。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/images//ex2.png" style="width: 600px; height: 600px;padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>输入：</strong>a = 5, b = 3, c = 3, d = 4, e = 5, f = 2
+<strong>输出：</strong>1
+<strong>解释：</strong>可以通过以下任一方式移动 1 次捕获黑皇后：
+- 将白色车移动到 (5, 2) 。
+- 将白色象移动到 (5, 2) 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= a, b, c, d, e, f &lt;= 8</code></li>
+	<li>两枚棋子不会同时出现在同一个格子上。</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/10000-10099/10036.Minimum Moves to Capture The Queen/README_EN.md

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+# [10036. Minimum Moves to Capture The Queen](https://leetcode.com/problems/minimum-moves-to-capture-the-queen)
+
+[中文文档](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README.md)
+
+## Description
+
+<p>There is a <strong>1-indexed</strong> <code>8 x 8</code> chessboard containing <code>3</code> pieces.</p>
+
+<p>You are given <code>6</code> integers <code>a</code>, <code>b</code>, <code>c</code>, <code>d</code>, <code>e</code>, and <code>f</code> where:</p>
+
+<ul>
+	<li><code>(a, b)</code> denotes the position of the white rook.</li>
+	<li><code>(c, d)</code> denotes the position of the white bishop.</li>
+	<li><code>(e, f)</code> denotes the position of the black queen.</li>
+</ul>
+
+<p>Given that you can only move the white pieces, return <em>the <strong>minimum</strong> number of moves required to capture the black queen</em>.</p>
+
+<p><strong>Note</strong> that:</p>
+
+<ul>
+	<li>Rooks can move any number of squares either vertically or horizontally, but cannot jump over other pieces.</li>
+	<li>Bishops can move any number of squares diagonally, but cannot jump over other pieces.</li>
+	<li>A rook or a bishop can capture the queen if it is located in a square that they can move to.</li>
+	<li>The queen does not move.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/images//ex1.png" style="width: 600px; height: 600px; padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>Input:</strong> a = 1, b = 1, c = 8, d = 8, e = 2, f = 3
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> We can capture the black queen in two moves by moving the white rook to (1, 3) then to (2, 3).
+It is impossible to capture the black queen in less than two moves since it is not being attacked by any of the pieces at the beginning.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/images//ex2.png" style="width: 600px; height: 600px;padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>Input:</strong> a = 5, b = 3, c = 3, d = 4, e = 5, f = 2
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> We can capture the black queen in a single move by doing one of the following: 
+- Move the white rook to (5, 2).
+- Move the white bishop to (5, 2).
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= a, b, c, d, e, f &lt;= 8</code></li>
+	<li>No two pieces are on the same square.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->