LeetCode 922 : Sort Array By Parity II

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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.

Note:

1. 2 <= A.length <= 20000
2. A.length % 2 == 0
3. 0 <= A[i] <= 1000

本题想要将数组中的数与数组下标的奇偶性统一，本人采用了以空间换时间的方式，将数组中的奇数与偶数分别挑出，再依次插入。Accepted代码如下：

class Solution {
    public int[] sortArrayByParityII(int[] A) {
        int l = A.length;
        int[] B = new int[l / 2];
        int[] C = new int[l / 2];
        for (int i = 0, j = 0, k = 0; i < A.length; i++) {
            if (A[i] % 2 == 0) {
                B[j++] = A[i];
            } else {
                C[k++] = A[i];
            }
        }
        for (int i = 0, j = 0, k = 0; i < A.length; i++) {
            if (i % 2 == 0) {
                A[i] = B[j++];
            } else {
                A[i] = C[k++];
            }
        }
        return A;
    }
}