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. 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. 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. 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. Replace the array nums with newNums.

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
}