0189. Rotate Array.md

题目

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

Follow up:

• 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?

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 <= 2 * 10^4
• It's guaranteed that nums[i] fits in a 32 bit-signed integer.
• k >= 0

我的思路

把这题的数组看成一个环，每次都把当前位置的元素像前移动k次，循环len(nums)次即可（因为需要移动的次数一定为len(nums)）。但运行了一下后，遇到了问题：如果环的长度是4，k=2，这种情况的话，每次都是位置0、2两个元素在移动，位置为1、3的两个元素将保持不动。因此修改一下代码，增加一个start位，如果在len(nums)次循环内发生下标又回到了start的情况，那么说明遇到了该问题，这种情况下就把start+1，然后继续循环即可。

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        l = len(nums)
        k = k % l
        start = 0
        i = start
        s = nums[i]
        for j in range(l):
            t = nums[(i+k)%l]
            nums[(i+k)%l] = s
            s = t
            i += k
            if (i % l == start):
                start += 1
                i = start
                s = nums[i%l]
        return