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
