922. Sort Array By Parity II

922. Sort Array By Parity II (Easy)

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

 

Solution 1. Two Pointers

// OJ: https://leetcode.com/problems/sort-array-by-parity-ii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& A) {
        for (int i = 0, j = 1, N = A.size(); i < N && j < N; i += 2, j += 2) {
            while (A[i] % 2 == 0) i += 2;
            while (A[j] % 2 != 0) j += 2;
            if (i < N) swap(A[i], A[j]);
        }
        return A;
    }
};