922. Sort Array By Parity II

Problem:

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:

2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000

思路：

Solution (C++):

vector<int> sortArrayByParityII(vector<int>& A) {
    if (A.empty())  return vector<int>{};
    int n = A.size();
    for (int i = 0; i < n-1; ++i) {
        int j = i+1;
        while ((i%2) ^ (A[i]%2) && j < n) {
            swap(A[i], A[j]);
            ++j;
        }
    }
    return A;
}

性能：

Runtime: 112 ms??Memory Usage: 9.5 MB

思路：

Solution (C++):

性能：

Runtime: ms??Memory Usage: MB

922. Sort Array By Parity II

原文：https://www.cnblogs.com/dysjtu1995/p/12622647.html