905. Sort Array By Parity（python+cpp）_905. sort array by parity python-优快云博客

本文介绍了一种将数组中的所有偶数元素置于奇数元素前的方法，并提供了Python和C++实现。通过使用额外空间和不使用额外空间两种方式，展示了如何有效地完成这一任务。

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题目：

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]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Note:
1 <= A.length <= 5000
0 <= A[i] <= 5000

解释：

python代码：

class Solution(object):
    def sortArrayByParity(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        return [x for x in A if x%2==0]+[x for x in A if x%2]

c++代码：

class Solution {
public:
    vector<int> sortArrayByParity(vector<int>& A) {
        vector<int> odd;
        vector<int> even;
        for(int i=0;i<A.size();i++)
        {
            if(A[i]%2==1)
            {
                odd.push_back(A[i]);
            }
            else
            {
                even.push_back(A[i]);
            }
            
        }
        even.insert(even.end(),odd.begin(),odd.end());
        return even;
        
    }
};

用两个指针可以降低空间复杂度

class Solution {
public:
    vector<int> sortArrayByParity(vector<int>& A) {
        //不需要额外空间的做法
        int n=A.size();
        vector<int> result(n,0);
        int i=0,j=n-1;
        for(auto num:A)
        {
            if (num%2)
                result[j--]=num;
            else
                result[i++]=num;
        }
        return result;
    }
};