[LeetCode] 922. Sort Array By Parity II(C++)

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

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