324. Wiggle Sort II【M】【33】

最新推荐文章于 2025-12-09 14:04:42 发布

原创最新推荐文章于 2025-12-09 14:04:42 发布 · 355 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #leetcode题解 #python #排序

Python 同时被 3 个专栏收录

176 篇文章

订阅专栏

leetcode

175 篇文章

订阅专栏

算法

157 篇文章

订阅专栏

本文深入探讨了Wiggle Sort算法，通过将未排序数组重新排列为交替升序和降序的方式，实现了一种独特的排序策略。详细介绍了算法的实现过程，并附带Python代码示例，同时分析了该算法的时间复杂度和空间复杂度，最后讨论了其在实际应用中的潜在用途及改进方向。

Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....

Example:
(1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6].
(2) Given nums = [1, 3, 2, 2, 3, 1], one possible answer is [2, 3, 1, 3, 1, 2].

Note:
You may assume all input has valid answer.

Follow Up:
Can you do it in O(n) time and/or in-place with O(1) extra space?

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

排序，分成两半，小的倒着排，大的正着排

class Solution(object):
    def wiggleSort(self, nums):

        nums.sort()
        l = len(nums)

        b = l / 2

        res = [0] * l
        flag = l%2

        for i in xrange(0,l,2):
            pos =  l-i-2 + flag
            res[pos] = nums[i/2]
        for i in xrange(1,l,2):
            pos = l - ((1 + i)/2 )
            res[i] = nums[pos]
        nums[:] = res[:]