Python3解leetcode Rotate Array

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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)

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)

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