[Swift]LeetCode324. 擺動排序 II | Wiggle Sort II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號:山青詠芝(shanqingyongzhi)
➤博客園地址:山青詠芝(https://www.cnblogs.com/strengthen/
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:http://www.javashuo.com/article/p-rbjctkdg-ke.html 
➤若是連接不是山青詠芝的博客園地址,則多是爬取做者的文章。
➤原文已修改更新!強烈建議點擊原文地址閱讀!支持做者!支持原創!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html

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

Example 1:github

Input: 
Output: One possible answer is .nums = [1, 5, 1, 1, 6, 4][1, 4, 1, 5, 1, 6]

Example 2:數組

Input: 
Output: One possible answer is .nums = [1, 3, 2, 2, 3, 1][2, 3, 1, 3, 1, 2]

Note:
You may assume all input has valid answer.微信

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


給定一個無序的數組 nums,將它從新排列成 nums[0] < nums[1] > nums[2] < nums[3]... 的順序。code

示例 1:htm

輸入: 
輸出: 一個可能的答案是 nums = [1, 5, 1, 1, 6, 4][1, 4, 1, 5, 1, 6]

示例 2:blog

輸入: 
輸出: 一個可能的答案是 nums = [1, 3, 2, 2, 3, 1][2, 3, 1, 3, 1, 2]

說明:
你能夠假設全部輸入都會獲得有效的結果。排序

進階:
你能用 O(n) 時間複雜度和 / 或原地 O(1) 額外空間來實現嗎?


380ms

 1 class Solution {
 2     func wiggleSort(_ nums: inout [Int]) {
 3         
 4         if nums.count < 2 {
 5             return
 6         }
 7         
 8         var count = nums.count
 9         
10         var begin = 0, end = count - 1, mid = (begin + end) >> 1
11         
12         func patition(_ begin : Int, _ end : Int) -> Int {
13             let index = nums[begin]
14             var low = begin , high = end
15             while low < high {
16                 while low < high && nums[high] >= index {
17                     high -= 1
18                 }
19                 if low < high {
20                     nums[low] = nums[high]
21                     low += 1
22                 }
23                 
24                 while low < high && nums[low] <= index {
25                     low += 1
26                 }
27                 
28                 if low < high {
29                     nums[high] = nums[low]
30                     high -= 1
31                 }
32                 
33             }
34             
35             nums[low] = index
36             
37             return low
38         }
39         
40         var index = patition(begin, end)
41         
42         while index != mid {
43             if index < mid {
44                 begin = index + 1
45             }else {
46                 end = index - 1
47             }
48             
49             index = patition(begin, end)
50         }
51         
52         mid = nums[index]
53         
54         var n = count
55         
56         func A(_ i : Int) -> Int {
57             return (1+2*i)%(count|1)
58         }
59         var i = 0, j = 0, k = count - 1
60         while j <= k {
61             if nums[A(j)] > mid {
62                 nums.swapAt(A(i), A(j))
63                 i+=1
64                 j+=1
65             }else if nums[A(j)] < mid {
66                 nums.swapAt(A(j), A(k))
67                 k-=1
68             }else {
69                 j+=1
70             }
71         }
72     }
73 }
相關文章
相關標籤/搜索