# 2293. Min Max Game

Easy
You are given a 0-indexed integer array `nums` whose length is a power of `2`.
Apply the following algorithm on `nums`:
1. 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. 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. 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. 4.
Replace the array `nums` with `newNums`.
5. 5.
Repeat the entire process starting from step 1.
Return the last number that remains in `nums` after applying the algorithm.
Example 1:
Input: nums = [1,3,5,2,4,8,2,2]
Output:
1
Explanation:
The following arrays are the results of applying the algorithm repeatedly.
First: nums = [1,5,4,2]
Second: nums = [1,4]
Third: nums = [1]
1 is the last remaining number, so we return 1.
Example 2:
Input: nums = [3]
Output:
3
Explanation:
3 is already the last remaining number, so we return 3.
Constraints:
• `1 <= nums.length <= 1024`
• `1 <= nums[i] <= 109`
• `nums.length` is a power of `2`.

### 解題

func minMaxGame(nums []int) int {
if len(nums) == 1 { return nums[0] }
res := make([]int ,0)
for i:=0; i<len(nums)/2; i++ {
index := i * 2
if i % 2 == 0 {
res = append(res, min(nums[index], nums[index+1]))
} else {
res = append(res, max(nums[index], nums[index+1]))
}
}
return minMaxGame(res)
}
func min(a , b int) int {
if a < b { return a }
return b
}
func max(a , b int) int {
if a > b { return a }
return b
}