Rotate List
https://oj.leetcode.com/problems/rotate-list/
Rotate List
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.
题目：又移链表K步，要考虑k大于链表长度的情况，用count计算链表长度，用k%count更新移动步数。
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 # @param {ListNode} head 9 # @param {integer} k 10 # @return {ListNode} 11 def rotateRight(self, head, k): 12 if k==0: return head #注意‘==’ 13 if head==None: return head 14 dummy=ListNode(0) 15 dummy.next=head 16 p=dummy 17 count=0 18 while p.next: #计算链表的长度 19 p=p.next 20 count += 1 21 step=count-(k%count) #向右移动k步，若k大于链表长度，则移动步数是(k%count) 22 p.next=dummy.next 23 for i in range(step): 24 p=p.next #通过step找到p的初始位置，使得p.next即为所有需要移动的链表部分 25 head=p.next #把需移动的p.next赋给头结点，转移到前面 26 p.next=None #并令p.next=None,结尾 27 return head #返回此时转移后的head(为什么return dummy.next 会出错)
