[LeetCode] 905. Sort Array By Parity

最新推荐文章于 2025-02-12 22:45:00 发布

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最新推荐文章于 2025-02-12 22:45:00 发布

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原文链接：https://segmentfault.com/a/1190000017573484

本文介绍了一种将数组中所有偶数元素放置在奇数元素前面的算法实现。通过使用双指针技术，该算法有效地对数组进行了重新排序，确保了所有偶数元素位于数组的前部，而所有奇数元素位于后部。这种排序方式对于处理大量数据或需要快速访问偶数元素的场景特别有用。

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Problem

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 <= 5000
0 <= A[i] <= 5000

Solution

class Solution {
    public int[] sortArrayByParity(int[] A) {
        if (A == null || A.length < 2) return A;
        int i = 0, j = 0;
        while (j < A.length) {
            while (i < j && A[i]%2 == 0) i++;
            if (A[j]%2 == 0) {
                int temp = A[i];
                A[i] = A[j];
                A[j] = temp;
                i++;
            }
            j++;
        }
        return A;
    }
}