【python】leetcode 189. Rotate Array （easy）

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189. Rotate Array （easy）

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?

c/c++解

【待续】

python解

我能想到的最优：使用了双端队列collections.deque，

注意不能直接nums = list(collections.deque(nums,maxlen=len(nums)).rotate(k%len(nums)))

因为这里list()已经是新地址了，赋给nums，nums已经不是原来的list了，题目要求in-place，即原list上修改

class Solution:
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        dq = collections.deque(nums,maxlen=len(nums))
        dq.rotate(k%len(nums))
        nums.clear()
        nums.extend(list(dq))

 Runtime: 48 ms, faster than 99.82% of Python3 

1 普通思路：末尾数字insert到开头，并删除末尾

但是这个方法有点慢：Runtime: 344 ms, faster than 0.98%

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.
        """
        for i in range(k): 
            nums.insert(0,nums[-1])
            del nums[-1]

2 python切片

Runtime: 68 ms, faster than 33.56% 

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.
        """
   
        length = len(nums) 
        nums.extend(nums[0:(length- k%length)])
        del nums[0:(length - k%length)]