LeetCode 61. 旋转链表

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。



 



示例 1：



输入：head = [1,2,3,4,5], k = 2

输出：[4,5,1,2,3]





示例 2：



输入：head = [0,1,2], k = 4

输出：[2,0,1]





 



提示：



链表中节点的数目在范围 [0, 500] 内

-100 <= Node.val <= 100

0 <= k <= 2 \* 109

# Definition for singly-linked list.

# class ListNode:

#     def \_\_init\_\_(self, val=0, next=None):

#         self.val = val

#         self.next = next

class Solution:

    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:

        if head  == None:

            return head

        headPoint = head

        length = 1

        while head.next != None:

            head = head.next

            length += 1

        endPoint = head

        endPoint.next = headPoint #形成一个循环

        moveLength = length - (k % length) - 1 #得出移动步数

        print(moveLength)

        while moveLength != 0:

            headPoint = headPoint.next

            moveLength -= 1

        endPoint = headPoint

        headPoint = headPoint.next

        endPoint.next = None

        return headPoint