動機

題目很賤，但賤的心服口服

reverse原來這麼泛用嗎?!

Problem

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 = 3Output: [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 = 2Output: [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?

Sol

如果老實的一個一個rotate，就會TLE

一個是Cyclic replace，就一直往前跳，直到再次回到起點

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        k = k % len(nums)
        changed,i = 0,0
        while changed < len(nums):
            start,source = i, nums[i]
            while True:
                nextI = (start+k) % len(nums)
                nums[nextI], source = source, nums[nextI]
                start, changed = nextI, changed+1
                if start == i:
                    break
            i += 1

還有另一個是reverse

先整個reverse，再reverse要rotate的個數

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        k = k % len(nums)
        def rev(ns,i,j):
            while i < j:
                ns[i], ns[j-1] = ns[j-1], ns[i]
                i,j = i+1, j-1
        
        rev(nums,0,len(nums))
        rev(nums,0,k)
        rev(nums,k,len(nums))