922. Sort Array By Parity II

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本文介绍了一种数组排序算法，该算法确保数组中所有偶数位于偶数索引位置，所有奇数位于奇数索引位置。通过遍历数组并交换不符合条件的元素来实现。示例输入[4,2,5,7]，输出可能为[4,5,2,7]，满足条件的其他输出也接受。

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Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

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

You may return any answer array that satisfies this condition.

Example 1:

Input: [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.

#pragma once
#include<iostream>
#include<vector>

using namespace std;


class Solution {
public:
	vector<int> sortArrayByParityII(vector<int>& A) {

		for (int i = 0; i < A.size(); i++)
		{
            //判断偶数位置上的数是不是偶数，不是则寻找后面为偶数的数，并替换
			if (i % 2 == 0&&A[i]%2!=0) 
			{
				for (int j = i + 1; j < A.size() ; j++)
				{
					if (A[j] % 2 == 0)
					{
						int temp = A[j];
						A[j] = A[i];
						A[i] = temp;
						break;
					}
				}
			}
            //判断偶数位置上的数是不是奇数，不是则寻找后面为奇数的数，并替换
            else if (i % 2 != 0 && A[i] % 2 == 0)
			{
				for (int j = i + 1; j < A.size(); j++)
				{
					if (A[j] % 2 != 0)
					{
						int temp = A[j];
						A[j] = A[i];
						A[i] = temp;
						break;
					}
				}
			}
			
		}
		return A;

	}
};