{ "data": { "question": { "questionId": "2386", "questionFrontendId": "2293", "categoryTitle": "Algorithms", "boundTopicId": 1532651, "title": "Min Max Game", "titleSlug": "min-max-game", "content": "

You are given a 0-indexed integer array nums whose length is a power of 2.

\n\n

Apply the following algorithm on nums:

\n\n
    \n\t
  1. Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.
    2. \n\t
  2. For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).
    3. \n\t
  3. For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).
    4. \n\t
  4. Replace the array nums with newNums.
    5. \n\t
  5. Repeat the entire process starting from step 1.
    6. \n
\n\n

Return the last number that remains in nums after applying the algorithm.

\n\n

 

\n

Example 1:

\n\"\"\n
\nInput: nums = [1,3,5,2,4,8,2,2]\nOutput: 1\nExplanation: 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
\n\n

Example 2:

\n\n
\nInput: nums = [3]\nOutput: 3\nExplanation: 3 is already the last remaining number, so we return 3.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "极大极小游戏", "translatedContent": "

给你一个下标从 0 开始的整数数组 nums ，其长度是 2 的幂。

\n\n

对 nums 执行下述算法：

\n\n
    \n\t
  1. 设 n 等于 nums 的长度，如果 n == 1 ，终止 算法过程。否则，创建 一个新的整数数组 newNums ，新数组长度为 n / 2 ，下标从 0 开始。
    2. \n\t
  2. 对于满足 0 <= i < n / 2 的每个 偶数 下标 i ，将 newNums[i] 赋值 为 min(nums[2 * i], nums[2 * i + 1]) 。
    3. \n\t
  3. 对于满足 0 <= i < n / 2 的每个 奇数 下标 i ，将 newNums[i] 赋值 为 max(nums[2 * i], nums[2 * i + 1]) 。
    4. \n\t
  4. 用 newNums 替换 nums 。
    5. \n\t
  5. 从步骤 1 开始 重复 整个过程。
    6. \n
\n\n

执行算法后，返回 nums 中剩下的那个数字。

\n\n

 

\n\n

示例 1：

\n\n

\"\"

\n\n
\n输入：nums = [1,3,5,2,4,8,2,2]\n输出：1\n解释：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
\n\n

示例 2：

\n\n
\n输入：nums = [3]\n输出：3\n解释：3 就是最后剩下的数字，返回 3 。\n
\n\n

 

\n\n

提示：

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