原题地址：https://oj.leetcode.com/problems/rotate-list/

题意：

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:

Given 1->2->3->4->5->NULL and k = 2,

return 4->5->1->2->3->NULL.

解题思路：循环右移一条链表，比如k=2，（1，2，3，4，5）循环右移两位变为（4，5，1，2，3）。由于k值有可能比链表长度大很多，所以先要用一个count变量求出链表的长度。而k%count就是循环右移的步数。

代码：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param head, a ListNode    # @param k, an integer    # @return a ListNode    def rotateRight(self, head, k):        if k == 0:            return head        if head == None:            return head        dummy = ListNode(0)        dummy.next = head        p = dummy        count = 0        while p.next:            p = p.next            count += 1        p.next = dummy.next        step = count - ( k % count )        for i in range(0, step):            p = p.next        head = p.next        p.next = None        return head