Easy

序列含有n个元素，返回往右旋转k步的序列

For example
若n = 7, k = 3,
序列 [1,2,3,4,5,6,7]旋转为 [5,6,7,1,2,3,4].

注意k可能大于n即可。

class Solution(object):
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        k = k % n
        nums[:] = nums[n-k:]+nums[:n-k]