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
