★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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 }