问题描述:

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?

思路:

解题思路比较多,最关键的想出尽可能多的解题方法

代码

class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
while k >= len(nums):
k -= len(nums)
if k == 0:
return
num1 = list([int])
num1[:] = nums[:]
nums[0:k] = num1[-k:]#利用num的后k个数字,替换nums的前k个数字
nums[k:] = num1[0:len(num1)-k]
nums[:] = nums[0:len(num1)]

以上代码,时间复杂度为O(1)

Runtime: 48 ms, faster than 94.32% of Python3 online submissions forRotate Array.
Memory Usage: 13.7 MB, less than 5.23% of Python3 online submissions for Rotate Array.
 
 
将代码优化:
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
k = k % len(nums)
nums[:] = nums[-k:] + nums[:-k]

以上代码:

Runtime: 48 ms, faster than 94.32% of Python3 online submissions forRotate Array.
Memory Usage: 13.5 MB, less than 32.26% of Python3 online submissions for Rotate Array.
 
以上两组代码相比,运算时间上没有太大的变化;第二个代码少用一个数组的空间,导致空间占用会比较少一些
 
 

class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
while k >= len(nums):
k -= len(nums)
if k == 0:
return
for i in range(k):
nums.insert(0,nums[-1])
nums.pop()

以上代码,时间复杂度O(n)

Runtime: 124 ms, faster than 16.90% of Python3 online submissions forRotate Array.
Memory Usage: 13.4 MB, less than 58.82% of Python3 online submissions for Rotate Array.

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