title

problem link

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.

solution

even elements first, followed by all the odd elements, could use two pointers, or use the new array record the result

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int[] ans = new int[A.length];
        int t = 0;

        for (int i = 0; i < A.length; ++i)
            if (A[i] % 2 == 0)
                ans[t++] = A[i];

        for (int i = 0; i < A.length; ++i)
            if (A[i] % 2 == 1)
                ans[t++] = A[i];

        return ans;
    }
}