【leetcode】61. Rotate List

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