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905. Sort Array By Parity - Easy

时间:2020-08-25 14:43:43      阅读:64      评论:0      收藏:0      [点我收藏+]

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. 1 <= A.length <= 5000
  2. 0 <= A[i] <= 5000

 

time = O(n), space = O(1)

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int i = 0, j = A.length - 1;
        while(i < j) {
            if(A[i] % 2 != 0 && A[j] % 2 == 0) {
                swap(A, i++, j--);
            }
            if(A[i] % 2 == 0) {
                i++;
            }
            if(A[j] % 2 != 0) {
                j--;
            }
        }
        return A;
    }
    
    public void swap(int[] A, int i, int j) {
        int tmp = A[i];
        A[i] = A[j];
        A[j] = tmp;
    }
}

 

905. Sort Array By Parity - Easy

原文:https://www.cnblogs.com/fatttcat/p/13559036.html

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