当前位置: 首页> 教育> 大学 > 苏州刚刚发生新闻_一个人免费看直播_软文推广网_下载谷歌浏览器

苏州刚刚发生新闻_一个人免费看直播_软文推广网_下载谷歌浏览器

时间:2025/7/10 3:36:56来源:https://blog.csdn.net/weixin_57266891/article/details/142727076 浏览次数:0次
苏州刚刚发生新闻_一个人免费看直播_软文推广网_下载谷歌浏览器

题目
Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]
Example 2:

Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Constraints:

1 <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
0 <= k <= 105

Follow up:

Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
Could you do it in-place with O(1) extra space?

解题思路
题目要求将array向右rotate k步。
我们的思路是先将整个array进行reverse,再将前k个进行reverse,最后将剩余的进行reverse。

例如,nums = [1, 2, 3, 4, 5, 6, 7]和k = 3。
reverse整个array,[7, 6, 5, 4, 3, 2, 1]。
reverse前3个elements,[5, 6, 7, 4, 3, 2, 1]。
reverse剩余的elements,[5, 6, 7, 1, 2, 3, 4]。

space complexity等于O(1)。

class Solution(object):def rotate(self, nums, k):""":type nums: List[int]:type k: int:rtype: None Do not return anything, modify nums in-place instead."""n = len(nums)k = k%n# Helper function to reverse elements in the array from index 'start' to 'end'def reverse(start, end):while start < end:nums[start], nums[end] = nums[end], nums[start]start += 1end -= 1reverse(0, n-1)reverse(0, k-1)reverse(k, n-1)return nums
关键字:苏州刚刚发生新闻_一个人免费看直播_软文推广网_下载谷歌浏览器

版权声明:

本网仅为发布的内容提供存储空间,不对发表、转载的内容提供任何形式的保证。凡本网注明“来源:XXX网络”的作品,均转载自其它媒体,著作权归作者所有,商业转载请联系作者获得授权,非商业转载请注明出处。

我们尊重并感谢每一位作者,均已注明文章来源和作者。如因作品内容、版权或其它问题,请及时与我们联系,联系邮箱:809451989@qq.com,投稿邮箱:809451989@qq.com

责任编辑: