Given an array, rotate the array to the right by k steps, where k is non-negative. 给定一个数组，将数组向右旋转k步，其中k为非负数。
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? 尝试尽可能多地提出解决方案，至少有3种不同的方法可以解决这个问题。 你能用O（1）额外空间就地做到吗？
思路：
- 使用额外的空间, 一个用队列进行存储，另一个使用栈进行存储
- 使用字符串的截取功能，将后面的字符串直接拼接到原字符串前面
- 也等价于将头部加到字符串的尾部
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.
"""
l = nums[:len(nums) - k]
l2 = nums[len(nums) - k:len(nums)]
nums = l2 + l
print(nums)
# print(l)
if __name__ == "__main__":
n = [1, 2, 3, 4, 5, 6, 7]
k = 3
Solution().rotate(n, k)
此方法出错，原因不知 — layout: post title: “189. Rotate Array” tag: leetcode刷题笔记 —
网上思路： https://blog.csdn.net/coder_orz/article/details/52052767
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.
"""
#重新赋值
k = k % len(nums)
nums[:k] = nums[len(nums)-k:]
nums[k:] = nums[:len(nums) - k]
print(nums)
if __name__ == "__main__":
n = [1, 2, 3, 4, 5, 6, 7]
k = 3
Solution().rotate(n, k)
