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
解题思路：
闭合为环
力扣
【Leetcode】Python&C++：61. 旋转链表 (链表操作)【每日一题系列20210327】_哔哩哔哩_bilibili
Python代码：
# 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 k == 0 or not head or not head.next:
return head
cur = head
n = 1
while cur.next:
cur = cur.next
n += 1
add = n - k % n
if add == n:
return head
cur.next = head
while add:
cur = cur.next
add -= 1
res = cur.next
cur.next = None
return res
