leetcode 【 Reverse Nodes in k-Group 】 python 实现

原题：

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

代码：oj测试通过 Runtime: 125 ms

 # 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 reverseKGroup(self, head, k):
         if head is None or head.next is None or k<2:
             return head
 
         dummyhead = ListNode(0)
         dummyhead.next = head
 
         slow = dummyhead
         fast = dummyhead
 
         while 1:
             sign = 0
             for i in range(k):
                 if fast.next is not None:
                     fast = fast.next
                 else:
                     sign = 1
                     break
             if sign == 1 :
                 break
             else:
                 curr = slow.next
                 for i in range(k-1):
                     tmp = curr.next
                     curr.next = tmp.next
                     tmp.next = slow.next
                     slow.next = tmp
                 slow = curr
                 fast = curr
 
         return dummyhead.next

思路：

这道题看主要有两个点：

1. 快慢指针技巧：fast指针在前面探路，如果不满足翻转的条件，退出直接返回；slow指针在后面跟着，slow.next始终指向待翻转的第一个ListNode位置

2. 链表翻转：这个我在之前的一篇日志中已经说明了，用了一个书本倒叙的例子，详情见http://www.cnblogs.com/xbf9xbf/p/4212159.html这篇日志。

总的思路就是把看似复杂的任务分解成小任务，然后逐个击破。