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] <= 5000

Solution

Time complexity : O(n)
Space complexity : O(1)

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

使用2-pointer的方式，從頭找到奇數，從尾找到偶數，兩者交換。