leetcode 【 Reorder List 】python 实现

题目:

Given a singly linked list LL0→L1→…→Ln-1→Ln,
reorder it to: L0→LnL1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

 

代码: oj 测试通过 248 ms

 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 head, a ListNode
 9     # @return nothing
10     def reorderList(self, head):
11         if head is None or head.next is None or head.next.next is None:
12             return head
13         
14         dummyhead = ListNode(0)
15         dummyhead.next = head
16         
17         # get the length of the linked list
18         p = head
19         list_length = 0
20         while p is not None:
21             list_length += 1
22             p = p.next
23         
24         #reverse the second half linked list
25         fast = dummyhead
26         for i in range((list_length+1)/2):
27             fast = fast.next
28         pre = fast
29         curr = pre.next
30         for i in range( (list_length)/2 - 1 ):
31             tmp = curr.next
32             curr.next = tmp.next
33             tmp.next = pre.next
34             pre.next = tmp
35         
36         #merge
37         h2 = pre.next
38         fast.next = None # cut the connection between 1st half linked list and 2nd half linked list
39         while head is not None and h2 is not None:
40             tmp = head.next
41             head.next = h2
42             tmp2 = h2.next
43             head.next.next = tmp
44             h2 = tmp2
45             head = tmp
46         
47         return dummyhead.next

思路

这道题的路子分三块:

1. 遍历单链表 求链表长度

2. 锁定后半个链表,反转后半个链表的每个元素

3. 切断前后半个链表的链接处 然后合并两个链表

posted on 2015-01-09 23:19  承续缘  阅读(275)  评论(1编辑  收藏  举报

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