leetcode-922-Sort-Array-By-Parity-II

发表于   |  

描述

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:

1
2
3
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

分析

定义一个 i 变量用来记录偶数位的索引，j 变量记录奇数位的索引，当 i, j 位置上的值都不匹配时，交换两值，时间复杂度是 $O(n)$。

解决方案1(Java)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution {
    public int[] sortArrayByParityII(int[] A) {
        for (int i = 0, j = 1; i < A.length && j < A.length; ) {
            if (A[i]%2 == 1 && A[j]%2 == 0) {
                A[i] = A[i] + A[j];
                A[j] = A[i] - A[j];
                A[i] = A[i] - A[j];
            }
            if (A[i]%2 == 0) {
                i += 2;
            }
            if (A[j]%2 == 1) {
                j += 2;
            }
        }
        return A;
    }
}

题目来源