One question daily 2293. Great mini game

Although it is a simple question，But there are recursive ideas。

Constantly convert the one -dimensional list into a two -dimensional list for operation（This can be avoidedindexChaos caused by changes）

Then useflagCount，Comparison of maximum and minimum values。

Just return to the end，Although it must be returned to a number，Still‘nums[0]’To avoid warning。

2293. Great mini game
Simple
39
company
Google Google
Give you a bidding 0 Started integer array nums ，Its length is 2 Power。

right nums Execute the following algorithm：

set up n equal nums length，if n == 1 ，termination Algorithm。otherwise，create A new integer array newNums ，The length of the new array is n / 2 ，Bidding from 0 start。
right于满足 0 <= i < n / 2 Every even Bidding i ，Will newNums[i] Assignment for min(nums[2 * i], nums[2 * i + 1]) 。
right于满足 0 <= i < n / 2 Every odd number Bidding i ，Will newNums[i] Assignment for max(nums[2 * i], nums[2 * i + 1]) 。
use newNums replace nums 。
From steps 1 start repeat the whole process。
After executing the algorithm，return nums The remaining numbers。

 

Exemplary example 1：



enter：nums = [1,3,5,2,4,8,2,2]
Output：1
explain：repeat执行算法会得到下述数组。
first round：nums = [1,5,4,2]
second round：nums = [1,4]
Third round：nums = [1]
1 Is the last number left，return 1 。
Exemplary example 2：

enter：nums = [3]
Output：3
explain：3 Is the last number left，return 3 。
 

hint：

1 <= nums.length <= 1024
1 <= nums[i] <= 109
nums.length yes 2 Power

class Solution:
    def minMaxGame(self, nums: List[int]) -> int:
        flag = 0
        while len(nums) > 1:
            nums = [nums[i:i + 2] for i in range(0, len(nums), 2)]
            # Divide2dimension
            for i in range(len(nums)):
                if flag % 2 == 0:
                    nums[i] = min(nums[i])
                    flag += 1
                else:
                    nums[i] = max(nums[i])
                    flag += 1
        return nums[0]