2164. Sort Even and Odd Indices Independently

# 2164. Sort Even and Odd Indices Independently#

## 题目 #

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`

## 题目大意 #

1. 按 非递增 顺序排列 nums 奇数下标 上的所有值。 举个例子，如果排序前 nums = [4,1,2,3] ，对奇数下标的值排序后变为 [4,3,2,1] 。奇数下标 1 和 3 的值按照非递增顺序重排。
2. 按 非递减 顺序排列 nums 偶数下标 上的所有值。 举个例子，如果排序前 nums = [4,1,2,3] ，对偶数下标的值排序后变为 [2,1,4,3] 。偶数下标 0 和 2 的值按照非递减顺序重排。

## 解题思路 #

• 简单题。分别将奇数和偶数位上的数字排序，奇数位的数从大到小，偶数位的数从小到大。最后将他们组合成一个数组。

## 代码 #

``````package leetcode

import (
"sort"
)

func sortEvenOdd(nums []int) []int {
odd, even, res := []int{}, []int{}, []int{}
for index, v := range nums {
if index%2 == 0 {
even = append(even, v)
} else {
odd = append(odd, v)
}
}
sort.Ints(even)
sort.Sort(sort.Reverse(sort.IntSlice(odd)))

indexO, indexE := 0, 0
for i := 0; i < len(nums); i++ {
if i%2 == 0 {
res = append(res, even[indexE])
indexE++
} else {
res = append(res, odd[indexO])
indexO++
}
}
return res
}
``````

Apr 8, 2023