[leetcode] 905. Sort Array By Parity

Description

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 <= 50000 <= A[i] <= 5000

分析

题目的意思是：这是一道easy的题目，需要把偶数放在前面，奇数放在后面，我的方法就是首尾进行遍历。也accept了，看标准答案了之后，发现了另一种解法，非常的简洁，我这里也分享一下，用的是lambda表达式，真是厉害了

代码一

class Solution: def sortArrayByParity(self, A: List[int]) -> List[int]: n=len(A) i=0 j=n-1 while(i<j): while(i<j and A[i]%2==0): i+=1 while(i<j and A[j]%2==1): j-=1 t=A[i] A[i]=A[j] A[j]=t i+=1 j-=1 return A

代码二

class Solution: def sortArrayByParity(self, A: List[int]) -> List[int]: A.sort(key=lambda x:x%2) return A

参考文献

[LeetCode] Approach 1: Sort