原题地址：https://oj.leetcode.com/problems/reverse-nodes-in-k-group/
题意：
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example,
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
解题思路：稍微难一点的链表反转题目。需要将链表反转的函数单独写成一个函数，这样看起来会清晰一些。
代码：
# 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 reverse(self, start, end): newhead=ListNode(0); newhead.next=start while newhead.next!=end: tmp=start.next start.next=tmp.next tmp.next=newhead.next newhead.next=tmp return [end, start] def reverseKGroup(self, head, k): if head==None: return None nhead=ListNode(0); nhead.next=head; start=nhead while start.next: end=start for i in range(k-1): end=end.next if end.next==None: return nhead.next res=self.reverse(start.next, end.next) start.next=res[0] start=res[1] return nhead.next