题目如下：

解题思路:首先求出链表的长度length，然后令k = k%length，得出的k即为要rotated长度。从头开始遍历链表到第k个元素，将k的next指向none，原链表的尾节点指向头结点即可。

代码如下：

class Solution(object):    def rotateRight(self, head, k):        """        :type head: ListNode        :type k: int        :rtype: ListNode        """        if k == 0 or head == None:            return head        length = 0        node = head        tail = None        while node != None:            if node != None:                tail = node            length += 1            node = node.next        k = k % length        if k == 0:            return head        i = 1        node = head        while i + k < length:            node = node.next            i += 1        newHead = node.next        node.next = None        tail.next = head        return newHead