Wiggle Sort II

最新推荐文章于 2024-08-13 10:30:50 发布

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原文链接：http://www.cnblogs.com/amazingzoe/p/5743223.html

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#runtime

本文介绍了一种使数组形成波动排序的算法实现，即确保数组中每个元素满足nums[0]<nums[1]>nums[2]<nums[3]...的条件。通过先对数组进行排序，然后交替放置较大和较小的元素来达到这一目的。

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?

Analyse: space: O(n), time: O(nlogn).

Runtime: 140ms

 1 class Solution {
 2 public:
 3     void wiggleSort(vector<int>& nums) {
 4         // the number with odd index should be larger than others
 5         // find the larger part of the array, then insert the smaller and larger part correspondingly
 6         vector<int> temp = nums;
 7         sort(temp.begin(), temp.end());
 8         
 9         int n = nums.size(), k = (n + 1) / 2;
10         for(int i = 0; i < nums.size(); i++)
11             nums[i] = i & 1 ? temp[--n] : temp[--k];
12     }
13 };