【leetcode】905. Sort Array By Parity

思路

分配两个与A同长的数组，分别保持偶数和奇数；有个指针标记共有多少个偶数，根据标记将奇数数组全部拼接在偶数之后。

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int[] even = new int[A.length];
        int[] odd = new int[A.length];
        int indexE = 0;
        int indexO = 0;

        for (int i=0; i<A.length; i++) {
            if (A[i]%2==0) {
                even[indexE++] = A[i];
            } else
                odd[indexO++] = A[i];
        }

        for (int j=0; j<indexO; j++) {
            even[indexE++] = odd[j];
        }

        return even;
    }
}