卖萌的弱渣

I am stupid, I am hungry.

Wiggle Sort II

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?

Solution

O(nlog(n)) + O(n)

  1. 对原数组排序，得到排序后的辅助数组snums

  2. 对原数组的偶数位下标填充snums的末尾元素 (current biggest)

  3. 对原数组的奇数位下标填充snums的末尾元素

(Wiggle-Sort2.py) download
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class Solution(object):
    def wiggleSort(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        # O(nlog(n)) + O(n)
        sorted_list = sorted(nums)
        n = len(sorted_list)
        # fill odd position
        # sorted_list.pop will return the current larggest one
        for i in range(1,n,2):
            nums[i] = sorted_list.pop()
        # fill even position
        for i in range(0,n,2):
            nums[i] = sorted_list.pop()

O(n) + O(1)

Ideas