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?
LeetCode：链接
和剑指Offer_编程题34：左旋转字符串是一样的思想，但是要求更加严格。
左旋转字符串：
(1) 先将左边 n个字符串进行翻转
（2）再将右边剩余字符串进行翻转
（3）最后将整个字符串进行翻转
右旋转字符串：
(1) 先将整个字符串进行翻转
（2）再将左边 n个字符串进行翻转
（3）再将右边剩余字符串进行翻转
而且左移或者右移的位数是可以大于数组长度的，一定要求余数；其次，一定要在原数组上操作，不能切片。
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.
"""
'''要求余数 如果数组长度是8 右移10位 实际上就是移动完一圈之后回到原位 再右移两位'''
k = k % len(nums)
'''一定要在原数组上操作 不能用切片的形式'''
self.reverse(nums, 0, len(nums)-1)
self.reverse(nums, 0, k-1)
self.reverse(nums, k, len(nums)-1)
def reverse(self, nums, start, end):
while start < end:
nums[start], nums[end] = nums[end], nums[start]
start += 1
end -= 1