feat: add new leetcode problems

Author: yanglbme

solution/1800-1899/1822.Sign of the Product of an Array/README.md

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+# [1822. 数组元素积的符号](https://leetcode-cn.com/problems/sign-of-the-product-of-an-array)
+
+[English Version](/solution/1800-1899/1822.Sign%20of%20the%20Product%20of%20an%20Array/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>已知函数 <code>signFunc(x)</code> 将会根据 <code>x</code> 的正负返回特定值：</p>
+
+<ul>
+	<li>如果 <code>x</code> 是正数，返回 <code>1</code> 。</li>
+	<li>如果 <code>x</code> 是负数，返回 <code>-1</code> 。</li>
+	<li>如果 <code>x</code> 是等于 <code>0</code> ，返回 <code>0</code> 。</li>
+</ul>
+
+<p>给你一个整数数组 <code>nums</code> 。令 <code>product</code> 为数组 <code>nums</code> 中所有元素值的乘积。</p>
+
+<p>返回 <code>signFunc(product)</code> 。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [-1,-2,-3,-4,3,2,1]
+<strong>输出：</strong>1
+<strong>解释：</strong>数组中所有值的乘积是 144 ，且 signFunc(144) = 1
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,5,0,2,-3]
+<strong>输出：</strong>0
+<strong>解释：</strong>数组中所有值的乘积是 0 ，且 signFunc(0) = 0
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [-1,1,-1,1,-1]
+<strong>输出：</strong>-1
+<strong>解释：</strong>数组中所有值的乘积是 -1 ，且 signFunc(-1) = -1
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= nums.length <= 1000</code></li>
+	<li><code>-100 <= nums[i] <= 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
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1822.Sign of the Product of an Array/README_EN.md

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+# [1822. Sign of the Product of an Array](https://leetcode.com/problems/sign-of-the-product-of-an-array)
+
+[中文文档](/solution/1800-1899/1822.Sign%20of%20the%20Product%20of%20an%20Array/README.md)
+
+## Description
+
+<p>There is a function <code>signFunc(x)</code> that returns:</p>
+
+<ul>
+	<li><code>1</code> if <code>x</code> is positive.</li>
+	<li><code>-1</code> if <code>x</code> is negative.</li>
+	<li><code>0</code> if <code>x</code> is equal to <code>0</code>.</li>
+</ul>
+
+<p>You are given an integer array <code>nums</code>. Let <code>product</code> be the product of all values in the array <code>nums</code>.</p>
+
+<p>Return <code>signFunc(product)</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-1,-2,-3,-4,3,2,1]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The product of all values in the array is 144, and signFunc(144) = 1
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,5,0,2,-3]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> The product of all values in the array is 0, and signFunc(0) = 0
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-1,1,-1,1,-1]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> The product of all values in the array is -1, and signFunc(-1) = -1
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1823.Find the Winner of the Circular Game/README.md

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+# [1823. 找出游戏的获胜者](https://leetcode-cn.com/problems/find-the-winner-of-the-circular-game)
+
+[English Version](/solution/1800-1899/1823.Find%20the%20Winner%20of%20the%20Circular%20Game/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>共有 <code>n</code> 名小伙伴一起做游戏。小伙伴们围成一圈，按 <strong>顺时针顺序</strong> 从 <code>1</code> 到 <code>n</code> 编号。确切地说，从第 <code>i</code> 名小伙伴顺时针移动一位会到达第 <code>(i+1)</code> 名小伙伴的位置，其中 <code>1 <= i < n</code> ，从第 <code>n</code> 名小伙伴顺时针移动一位会回到第 <code>1</code> 名小伙伴的位置。</p>
+
+<p>游戏遵循如下规则：</p>
+
+<ol>
+	<li>从第 <code>1</code> 名小伙伴所在位置 <strong>开始</strong> 。</li>
+	<li>沿着顺时针方向数 <code>k</code> 名小伙伴，计数时需要 <strong>包含</strong> 起始时的那位小伙伴。逐个绕圈进行计数，一些小伙伴可能会被数过不止一次。</li>
+	<li>你数到的最后一名小伙伴需要离开圈子，并视作输掉游戏。</li>
+	<li>如果圈子中仍然有不止一名小伙伴，从刚刚输掉的小伙伴的 <strong>顺时针下一位</strong> 小伙伴 <strong>开始</strong>，回到步骤 <code>2</code> 继续执行。</li>
+	<li>否则，圈子中最后一名小伙伴赢得游戏。</li>
+</ol>
+
+<p>给你参与游戏的小伙伴总数 <code>n</code> ，和一个整数 <code>k</code> ，返回游戏的获胜者。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+![](./images/ic234-q2-ex11.png)
+
+<pre>
+<strong>输入：</strong>n = 5, k = 2
+<strong>输出：</strong>3
+<strong>解释：</strong>游戏运行步骤如下：
+1) 从小伙伴 1 开始。
+2) 顺时针数 2 名小伙伴，也就是小伙伴 1 和 2 。
+3) 小伙伴 2 离开圈子。下一次从小伙伴 3 开始。
+4) 顺时针数 2 名小伙伴，也就是小伙伴 3 和 4 。
+5) 小伙伴 4 离开圈子。下一次从小伙伴 5 开始。
+6) 顺时针数 2 名小伙伴，也就是小伙伴 5 和 1 。
+7) 小伙伴 1 离开圈子。下一次从小伙伴 3 开始。
+8) 顺时针数 2 名小伙伴，也就是小伙伴 3 和 5 。
+9) 小伙伴 5 离开圈子。只剩下小伙伴 3 。所以小伙伴 3 是游戏的获胜者。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 6, k = 5
+<strong>输出：</strong>1
+<strong>解释：</strong>小伙伴离开圈子的顺序：5、4、6、2、3 。小伙伴 1 是游戏的获胜者。
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= k <= n <= 500</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
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1823.Find the Winner of the Circular Game/README_EN.md

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+# [1823. Find the Winner of the Circular Game](https://leetcode.com/problems/find-the-winner-of-the-circular-game)
+
+[中文文档](/solution/1800-1899/1823.Find%20the%20Winner%20of%20the%20Circular%20Game/README.md)
+
+## Description
+
+<p>There are <code>n</code> friends that are playing a game. The friends are sitting in a circle and are numbered from <code>1</code> to <code>n</code> in <strong>clockwise order</strong>. More formally, moving clockwise from the <code>i<sup>th</sup></code> friend brings you to the <code>(i+1)<sup>th</sup></code> friend for <code>1 &lt;= i &lt; n</code>, and moving clockwise from the <code>n<sup>th</sup></code> friend brings you to the <code>1<sup>st</sup></code> friend.</p>
+
+<p>The rules of the game are as follows:</p>
+
+<ol>
+	<li><strong>Start</strong> at the <code>1<sup>st</sup></code> friend.</li>
+	<li>Count the next <code>k</code> friends in the clockwise direction <strong>including</strong> the friend you started at. The counting wraps around the circle and may count some friends more than once.</li>
+	<li>The last friend you counted leaves the circle and loses the game.</li>
+	<li>If there is still more than one friend in the circle, go back to step <code>2</code> <strong>starting</strong> from the friend <strong>immediately clockwise</strong> of the friend who just lost and repeat.</li>
+	<li>Else, the last friend in the circle wins the game.</li>
+</ol>
+
+<p>Given the number of friends, <code>n</code>, and an integer <code>k</code>, return <em>the winner of the game</em>.</p>
+
+<p>&nbsp;</p>
+
+<p><strong>Example 1:</strong></p>
+
+![](./images/ic234-q2-ex11.png)
+
+<pre>
+
+<strong>Input:</strong> n = 5, k = 2
+
+<strong>Output:</strong> 3
+
+<strong>Explanation:</strong> Here are the steps of the game:
+
+1) Start at friend 1.
+
+2) Count 2 friends clockwise, which are friends 1 and 2.
+
+3) Friend 2 leaves the circle. Next start is friend 3.
+
+4) Count 2 friends clockwise, which are friends 3 and 4.
+
+5) Friend 4 leaves the circle. Next start is friend 5.
+
+6) Count 2 friends clockwise, which are friends 5 and 1.
+
+7) Friend 1 leaves the circle. Next start is friend 3.
+
+8) Count 2 friends clockwise, which are friends 3 and 5.
+
+9) Friend 5 leaves the circle. Only friend 3 is left, so they are the winner.</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+
+<strong>Input:</strong> n = 6, k = 5
+
+<strong>Output:</strong> 1
+
+<strong>Explanation:</strong> The friends leave in this order: 5, 4, 6, 2, 3. The winner is friend 1.
+
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= n &lt;= 500</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->