leetcode-problemset/leetcode-cn/originData/min-max-game.json

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            "title": "Min Max Game",
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            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> whose length is a power of <code>2</code>.</p>\n\n<p>Apply the following algorithm on <code>nums</code>:</p>\n\n<ol>\n\t<li>Let <code>n</code> be the length of <code>nums</code>. If <code>n == 1</code>, <strong>end</strong> the process. Otherwise, <strong>create</strong> a new <strong>0-indexed</strong> integer array <code>newNums</code> of length <code>n / 2</code>.</li>\n\t<li>For every <strong>even</strong> index <code>i</code> where <code>0 &lt;= i &lt; n / 2</code>, <strong>assign</strong> the value of <code>newNums[i]</code> as <code>min(nums[2 * i], nums[2 * i + 1])</code>.</li>\n\t<li>For every <strong>odd</strong> index <code>i</code> where <code>0 &lt;= i &lt; n / 2</code>, <strong>assign</strong> the value of <code>newNums[i]</code> as <code>max(nums[2 * i], nums[2 * i + 1])</code>.</li>\n\t<li><strong>Replace</strong> the array <code>nums</code> with <code>newNums</code>.</li>\n\t<li><strong>Repeat</strong> the entire process starting from step 1.</li>\n</ol>\n\n<p>Return <em>the last number that remains in </em><code>nums</code><em> after applying the algorithm.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/04/13/example1drawio-1.png\" style=\"width: 500px; height: 240px;\" />\n<pre>\n<strong>Input:</strong> nums = [1,3,5,2,4,8,2,2]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The following arrays are the results of applying the algorithm repeatedly.\nFirst: nums = [1,5,4,2]\nSecond: nums = [1,4]\nThird: nums = [1]\n1 is the last remaining number, so we return 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> 3 is already the last remaining number, so we return 3.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1024</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>nums.length</code> is a power of <code>2</code>.</li>\n</ul>\n",
            "translatedTitle": "极大极小游戏",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，其长度是 <code>2</code> 的幂。</p>\n\n<p>对 <code>nums</code> 执行下述算法：</p>\n\n<ol>\n\t<li>设 <code>n</code> 等于 <code>nums</code> 的长度，如果 <code>n == 1</code> ，<strong>终止</strong> 算法过程。否则，<strong>创建</strong> 一个新的整数数组&nbsp;<code>newNums</code> ，新数组长度为 <code>n / 2</code> ，下标从 <strong>0</strong> 开始。</li>\n\t<li>对于满足&nbsp;<code>0 &lt;= i &lt; n / 2</code> 的每个 <strong>偶数</strong> 下标 <code>i</code> ，将 <code>newNums[i]</code> <strong>赋值</strong> 为 <code>min(nums[2 * i], nums[2 * i + 1])</code> 。</li>\n\t<li>对于满足&nbsp;<code>0 &lt;= i &lt; n / 2</code> 的每个 <strong>奇数</strong> 下标 <code>i</code> ，将 <code>newNums[i]</code> <strong>赋值</strong> 为 <code>max(nums[2 * i], nums[2 * i + 1])</code> 。</li>\n\t<li>用 <code>newNums</code> 替换 <code>nums</code> 。</li>\n\t<li>从步骤 1 开始 <strong>重复</strong> 整个过程。</li>\n</ol>\n\n<p>执行算法后，返回 <code>nums</code> 中剩下的那个数字。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/04/13/example1drawio-1.png\" style=\"width: 500px; height: 240px;\" /></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,3,5,2,4,8,2,2]\n<strong>输出：</strong>1\n<strong>解释：</strong>重复执行算法会得到下述数组。\n第一轮：nums = [1,5,4,2]\n第二轮：nums = [1,4]\n第三轮：nums = [1]\n1 是最后剩下的那个数字，返回 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3]\n<strong>输出：</strong>3\n<strong>解释：</strong>3 就是最后剩下的数字，返回 3 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1024</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>nums.length</code> 是 <code>2</code> 的幂</li>\n</ul>\n",
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