public class Solution extends Object
2164 - Sort Even and Odd Indices Independently.
Easy
You are given a 0-indexed integer array nums. Rearrange the values of nums according to the following rules:
nums in non-increasing order.
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.nums in non-decreasing order.
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 <= 1001 <= nums[i] <= 100| Constructor and Description |
|---|
Solution() |
| Modifier and Type | Method and Description |
|---|---|
int[] |
sortEvenOdd(int[] nums) |
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