Reverse Nodes in k-Group

 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 reverse(self,start,end):
12         newhead=ListNode(0);newhead.next=start
13         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.next
15             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=head
20         start=nhead
21         while start.next:
22             end=start
23             for i in range (k-1):                     #for i in range() 
24                 end=end.next
25                 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.next
27             res=self.reverse(start.next, end.next)
28             start.next=res[0]
29             start=res[1]
30         return nhead.next