[LeetCode&Python] Problem 905: Sort Array By Parity

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

Try:

In the beginning, I want to use recursive method to solve this problem.

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

        if not A: return []
        if A[0]%2!=0:
            return self.sortArrayByParity(A[1:])+[A[0]]
        return [A[0]]+self.sortArrayByParity(A[1:])

The problem is that this method needs too much memory space.

Solution:

Just use one line, you can solve this problem.

class Solution:
    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!=0]

原文地址：https://www.cnblogs.com/chiyeung/p/9655829.html