Rotate List Python Leetcode

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.

这题的做法是先遍历一遍把这个链表的头和尾联起来，同时计算总长度count，然后再走 count-k布。

把这里记为新的head 返回。

这里有点要注意，有时候k的值会远大于count所以要 dif=count-k%count

代码如下。# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None

class Solution:
# @param head, a ListNode
# @param k, an integer
# @return a ListNode
def rotateRight(self, head, k):
if not head or k==0:
return head
dummy=ListNode(0)
dummy.next=head
p=dummy
count=0
while p.next:
p=p.next
count+=1
p.next=dummy.next
dif=count-k%count
while dif>0:
p=p.next
dif-=1
head=p.next
p.next=None
return head