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
}