61.旋转链表

难度：中等
题目描述：

思路总结：思路呢，很简单，无非就两点，连环，双指针。最后需要求余n，以防k>>n的情况
题解一：
个人版-超时，改进几行代码后，AC。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        #思路：先将链表变成循环链表，然后在进行切割。
        if not head:return head
        cur = head
        n = 1
        while cur.next:
            cur = cur.next
            n += 1
        tail = cur
        cur.next = head
        fast = head
        for i in range(k%n):
            fast = fast.next
        while fast != cur:
            fast =fast.next
            head = head.next
        res = head.next
        head.next = None
        return res

题解一结果：

来源：CSDN

作者：LotusQ

链接：https://blog.csdn.net/qq_30057549/article/details/103788318