class Solution {
public:
    vector<int> sortArrayByParity(vector<int>& A) {
        int i = 0;int j = A.size() - 1;
        while(i < j){
            while(i < A.size() && A[i] % 2 == 0){
                i ++;
            }
            while(j >= 0 && A[j] % 2 == 1){
                j --;
            }
            if(i < j){
                int tmp = A[i];
                A[i] = A[j];
                A[j] = tmp;
                i++;j--;
            }
        }
        return A;
    }
};

905. Sort Array By Parity

Easy

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Given an array ​​A​​​ of non-negative integers, return an array consisting of all the even elements of ​​A​​​, followed by all the odd elements of ​​A​​.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4] Output: [2,4,3,1] The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Note:

1. ​​1 <= A.length <= 5000​​
2. ​​0 <= A[i] <= 500​​