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

最新推荐文章于 2025-08-15 08:27:29 发布

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最新推荐文章于 2025-08-15 08:27:29 发布

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文章标签： python

原文链接：http://www.cnblogs.com/chiyeung/p/9655829.html

本文介绍了一种简单有效的方法，用于将数组中的所有偶数元素放置在所有奇数元素之前。通过使用递归方法和更优的一行代码解决方案，文章详细解释了如何实现这一功能，并附带了一个Python代码示例。

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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 <= A.length <= 5000
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]

转载于:https://www.cnblogs.com/chiyeung/p/9655829.html