LeetCode周赛#106 Q1 Sort Array By Parity II

题目来源：https://leetcode.com/contest/weekly-contest-106/problems/sort-array-by-parity-ii/

问题描述

922. Sort Array By Parity II

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

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题意

给定一个长度为n（n是偶数）的数列，有n/2个元素是偶数，n/2个元素是奇数。求数列的一个重排，使得偶数下标的元素都是偶数，奇数下标的元素都是奇数。

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思路

设置两个指针i和j分别遍历数列的奇数下标和偶数下标，如果发现i下标位置和j下标位置下标与元素的奇偶性不同就交换。

以后千万要注意，“==”“!=”的优先级比位运算高！！！！！！！！

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代码

class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& A) {
        int i= 0 , j = 1, n = A.size();
        while (i<n && j<n)
        {
            while ((A[i] & 1) == 0)
            {
                i += 2;
            }
            if (i >= n)
            {
                break;
            }
            while ((A[j] & 1) == 1)
            {
                j += 2;
            }
            if (j >= n)
            {
                break;
            }
            swap(A[i], A[j]);
        }
        return A;
    }
};