189-rotate-array

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Description

Given an array, rotate the array to the right by k steps, where k is non-negative.

 

Example 1:

Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]

 

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

 

Follow up:

• Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Solution(Python)

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        self.usinglistcomprehension(nums, k)

    def usinglistcomprehension(self, nums: List[int], k: int) -> None:
        k %= len(nums)
        nums[:] = nums[-k:] + nums[:-k]

    def usingreverse(self, nums: List[int], k: int) -> None:
        k %= len(nums)
        self.reverse(nums, 0, len(nums) - 1)
        self.reverse(nums, 0, k - 1)
        self.reverse(nums, k, len(nums) - 1)

    def usingextraArray(self, nums: List[int], k: int) -> None:
        n = len(nums)
        tmp = [0] * n
        for i in range(n):
            tmp[(i + k) % len(nums)] = nums[i]
        for i in range(n):
            nums[i] = tmp[i]

    def bruteforce(self, nums: List[int], k: int) -> None:
        for i in range(k):
            prev = nums[-1]
            for j in range(len(nums)):
                nums[j], prev = prev, nums[j]

    def reverse(self, nums, start, end):
        while start < end:
            nums[start], nums[end] = nums[end], nums[start]
            start += 1
            end -= 1