Python3解leetcode Rotate Array
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.
"""
k = k % len(nums)
nums[:] = nums[-k:] + nums[:-k]
以上代码:
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
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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