leetcode讲解--922. Sort Array By Parity II

发布于 2018-12-25  约 3 分钟

题目

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

题目地址

讲解

这道题需要新开一个等大的空间,还需要两个指针来分别记录偶数和奇数的位置。

Java代码

class Solution {
    public int[] sortArrayByParityII(int[] A) {
        int[] result = new int[A.length];
        int indexOfEven = 0;
        int indexOfOdd = 1;
        for(int i=0;i<A.length;i++){
            if(A[i]%2==0){
                result[indexOfEven] = A[i];
                indexOfEven += 2;
            }else{
                result[indexOfOdd] = A[i];
                indexOfOdd += 2;
            }
        }
        return result;
    }
}
阅读 927发布于 2018-12-25

推荐阅读
liuqinh2s的专栏
用户专栏

我是一个有思想深度的人,所以博客里面会有大量的方法论,慎入

5 人关注
80 篇文章
专栏主页
目录