Given an array, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: [1,2,3,4,5,6,7] and 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: [-1,-100,3,99] and 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]
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Could you do it in-place with O(1) extra space?

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        k = k % n
        # 上一步是为了 Corner case: [1,2,3,4,5,6]   k = 11 的情况
        self.helper(nums,0, n-k-1)
        self.helper(nums,n-k,n-1)
        self.helper(nums,0,n-1)
    def helper(self,nums,start,end):
        while start < end:
            temp =  nums[start]
            nums[start] = nums[end]
            nums[end] = temp
            start += 1
            end -= 1
        # print(nums)
"""
知识点：
1、nums[0:n-k-1] 截取 list 片段是新生成一个 list，不改变原来的list
"""