Leetcode 2164. Sort Even and Odd Indices Independently

You are given a 0-indexed integer array nums. Rearrange the values of nums according to the following rules:

1. Sort the values at odd indices of nums in non-increasing order.

• For example, if nums = [4,1,2,3] before this step, it becomes [4,3,2,1] after. The values at odd indices 1 and 3 are sorted in non-increasing order.

2. Sort the values at even indices of nums in non-decreasing order.

• For example, if nums = [4,1,2,3] before this step, it becomes [2,1,4,3] after. The values at even indices 0 and 2 are sorted in non-decreasing order.

Return the array formed after rearranging the values of nums.

Example 1:

Input: nums = [4,1,2,3]

Output: [2,3,4,1]

Explanation: First, we sort the values present at odd indices (1 and 3) in non-increasing order. So, nums changes from [4,1,2,3] to [4,3,2,1]. Next, we sort the values present at even indices (0 and 2) in non-decreasing order. So, nums changes from [4,1,2,3] to [2,3,4,1]. Thus, the array formed after rearranging the values is [2,3,4,1].

Example 2:

Input: nums = [2,1]

Output: [2,1]

Explanation: Since there is exactly one odd index and one even index, no rearrangement of values takes place. The resultant array formed is [2,1], which is the same as the initial array.

Constraints:

• 1 <= nums.length <= 100

• 1 <= nums[i] <= 100

• easy 题目，运用list sort reverse函数即可

Runtime: 5 ms, faster than 42.52% of Java online submissions for Sort Even and Odd Indices Independently.

Memory Usage: 42.7 MB, less than 34.55% of Java online submissions for Sort Even and Odd Indices Independently.