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

方法1: partition + two pointers

思路：

首先沿用上一题的方法将偶数都放在前面，偶数在后。此时由于已知奇偶相等，可以用两个指针找到不前后都不在其位的两个数字，交换一下。这个方法和quick sort中的partition及其类似。

Complexity

Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& A) {
        int left = 0, right = 0, n = A.size();
        while (right < A.size()) {
            if (A[right] % 2 == 0) swap(A[left++], A[right++]);
            else right++;
        }
        
        left = 0, right = n / 2;
        while (right < n){
            if (left % 2 == 0) left++;
            else if (right % 2 == 1) right++;
            else {
                swap(A[left++], A[right++]);
            }
        }
        return A;
    }
};