chore: add daily leetcode post 2293_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/2293_translated.md

@@ -0,0 +1,81 @@
+---
+title: One question daily 2293. Great mini game
+date: '2024.01.01 0:00'
+tags:
+  - - Python
+  - - solved
+    - answer
+abbrlink: 9df6242c
+category:
+---
+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。
+
+
+```html
+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
+
+```
+
+```python
+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]
+
+```