一、题目

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

Example 1:

Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
Example 2:

Input: nums = [2,3]
Output: [2,3]

Constraints:

2 <= nums.length <= 2 * 104
nums.length is even.
Half of the integers in nums are even.
0 <= nums[i] <= 1000

Follow Up: Could you solve it in-place?

二、题解

class Solution {
public:vector<int> sortArrayByParityII(vector<int>& nums) {int n = nums.size();int oddIndex = 1;for(int i = 0;i < n;i+=2){//存在偶数位上的奇数if(nums[i] % 2 == 1){//查找奇数位上的偶数while(nums[oddIndex] % 2 != 0) oddIndex += 2;swap(nums[i],nums[oddIndex]);}}return nums;}
};