https://oj.leetcode.com/problems/reverse-nodes-in-k-group/

Reverse Nodes in k-Group

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

题目：k的倍数项反转，不足k倍数不变

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} k10     # @return {ListNode}11     def reverse(self,start,end):12         newhead=ListNode(0);newhead.next=start13         while newhead.next!=end:                      #把start后面的数依次放到最前面，直到end出现在最前面（头结点后）为止，实现逆转14             temp=start.next; start.next=temp.next15             temp.next=newhead.next; newhead.next=temp #不能写temp.next=start,(1,2,3)-(2,1,3)可以（2,1,3)-(3,2,1)错误16         return [end,start]17     def reverseKGroup(self, head, k):18         if head==None: return None                    #漏掉19         nhead=ListNode(0); nhead.next=head20         start=nhead21         while start.next:22             end=start23             for i in range (k-1):                     #for i in range() 24                 end=end.next25                 if end.next==None:                    #if在for循环里，走一步一判断，若在k-1步及之前为空，则不足k倍数，不反转直接输出26                     return nhead.next27             res=self.reverse(start.next, end.next)28             start.next=res[0]29             start=res[1]30         return nhead.next