Leetcode_Array --905. Sort Array By Parity [easy]

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.
给定一个非负整数数组A，请返回一个数组，这个数组的前面是A中的偶数，偶数后面跟着A中的奇数
You may return any answer array that satisfies this condition.
返回任意满足要求的数组即可

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

Solutions：

Python

class Solution:
    def sortArrayByParity(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """

        even,odd = [],[]
        for i in A:
            if i%2 == 0:
                even.append(i)
            else:
                odd.append(i)
        return even+odd

C++

class Solution {
public:
    vector sortArrayByParity(vector& A) {
        vector even;
        vector odd;
        for (int i : A){
            if (i%2 == 0){
                even.push_back(i);
            }
            else{
                odd.push_back(i);
            }
        }
        even.insert(even.end(),odd.begin(),odd.end());   #在vector even后面插入vector odd；even.insert(even.begin(),odd,begin(),odd.end())为在even前面插入odd
        return even;
        
    }
};