Given an unsorted array nums
,reorder it such that nums[0] < nums[1] > nums[2] < nums[3]...
.
Example 1:
Input: Output: One possible answer is .nums = [1,5,1,6,4][1,4,6]
Example 2:
Input: Output: One possible answer is .nums = [1,3,2,1][2,2]
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?
给定一个无序的数组 nums
,将它重新排列成 nums[0] < nums[1] > nums[2] < nums[3]...
的顺序。
示例 1:
输入: 输出: 一个可能的答案是 nums = [1,6]
示例 2:
输入: 输出: 一个可能的答案是 nums = [1,2]
说明:
你可以假设所有输入都会得到有效的结果。
进阶:
你能用 O(n) 时间复杂度和 / 或原地 O(1) 额外空间来实现吗?
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 }
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 [email protected] 举报,一经查实,本站将立刻删除。