Leetcode——Sort Array By Parity（905）（持续更新）

Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Note:

1. 1 <= A.length <= 5000
2. 0 <= A[i] <= 5000

分析：

将数组按照奇偶性分类，因为不要求顺序，用快排的partition函数即可。代码：

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

上面代码跑了28ms，应该不是最快的，因为是新题目，还没看到别人的思路，有了新方法会更新。