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。