LeetCode-Sort Array By Parity

来源：恒创科技编辑：恒创科技编辑部

2024-01-25 22:56:59

Description：
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 <= A.length <= 50000 <= A[i] <= 5000

题意：给定一个一维数组，将数组中的偶数移动到数组的首部，奇数放在数组的尾部；

解法：我们可以定义两个变量st和ed，st用于指示从首部开始的遇到的奇数，ed用于指示从尾部开始遇到的偶数，将这两个位置的元素进行交换，一直重复这个操作知道st > ed，这样所有的偶数就被交换到了首部，奇数被交换到了尾部；

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int st = 0;
        int ed = A.length - 1;
        while (st < ed) {
            while (st < A.length && A[st] % 2 == 0) st++;
            while (ed >= 0 && A[ed] % 2 == 1) ed--;
            if (st >= ed) break;
            A[st] = A[st] ^ A[ed];
            A[ed] = A[st] ^ A[ed];
            A[st] = A[st] ^ A[ed];
            st++;
            ed--;
        }
        return A;