Given an integer array nums, 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 - 10 <= 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)
```Python
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
self.inplaceoptimal(nums, k)
# poping each element and appending in front
# Time=complexity: O(n,k)
# Space complexity: O(n)
def bruteforce(self, nums, k):
k %= len(nums)
for _ in range(k):
nums.insert(0, nums.pop())
# use list com,phgrehension inspace
# Time=complexity: O(n)
# Space complexity: O(n)
def listcomprehension(self, nums, k):
k = k% len(nums)
nums[:] = nums[-k:] + nums[:-k]
# use inplace
# Time=complexity: O(n)
# Space complexity: O(1)
def inplaceoptimal(self, nums, k):
n = len(nums)
k %= n
def reverse(l ,r):
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l+=1
r -= 1
reverse(0,n -1 )
reverse(0, k -1)
reverse(k , n-1)
```