922. Sort Array By Parity II

最新推荐文章于 2024-07-12 16:48:32 发布

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最新推荐文章于 2024-07-12 16:48:32 发布

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原文链接：http://www.cnblogs.com/bernieloveslife/p/9797409.html

本文介绍了一种将数组中奇数和偶数分别放置在奇数位和偶数位的排序算法。通过创建两个列表分别存储奇数和偶数，然后交替地将它们放回原数组，以满足条件。示例输入[4,2,5,7]的一个可能输出是[4,5,2,7]。

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Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

Example 1:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Note:

2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000

class Solution:
    def sortArrayByParityII(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        odd = []
        even = []
        for i in A:
            if i%2==1:
                odd.append(i)
            else:
                even.append(i)
        pos = 0
        for i in range(len(odd)):
            A[pos] = even[i]
            A[pos+1] = odd[i]
            pos +=2
        return A

转载于:https://www.cnblogs.com/bernieloveslife/p/9797409.html