Problem

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

Example 1:

1
2
3
4
5
6
7
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:

1
2
3
4
5
6
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?

Solve

錯1

1
2
3
4
5
6
7
class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        nums = nums[ n - k:] + nums[:n-k]

由於leetcode 判定nums = nums[ n - k:n] + nums[0:n-k]

為建造新的一個變數，所以原先的並未改變

改成(但資測一樣錯的)，只是記錄一下leetcode的機制

nums=

→

nums[:]

1
2
3
4
5
6
7
8
9
10
11
12
13
class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        nums[:] = nums[ n - k:] + nums[:n-k]
'''
nums = [1,2]
k=5
Output = [1,2]
Expected = [2,1]
'''

解

1
2
3
4
5
6
7
8
class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        k = k % n
        nums[:] = nums[ n - k:n] + nums[0:n-k]

%做餘數

確保k 在[0,n-1]