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]