题目链接：Leetcode 61. Rotate List

本题采用快慢指针的思路，让fast指针比slow指针多走k步，当fast走到链表最后一个结点时，断开slow指针所在位置，使其指向None；并将fast指针指向链表头结点。

考虑到k值可能大于链表长度，需将k值在链表长度上进行取模运算。

完整代码如下：

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head: return head # base case

        fast, slow, l = head, head, head
        n = 0
        while l:
            n += 1
            l = l.next
        k %= n # 取模

        if k == 0: return head # 若k取模后等于0，那是否旋转链表并无差别，直接返回链表本身

        while k:
            fast = fast.next
            k -= 1
        while fast.next:
            slow, fast = slow.next, fast.next
        new_head = slow.next
        slow.next = None
        fast.next = head
        return new_head