leetcode 【 Intersection of Two Linked Lists 】python 实现

题目：

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

代码：oj在线测试通过 Runtime: 1604 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 two ListNodes
 9     # @return the intersected ListNode
10     def getListLen(self,head):
11         length = 0
12         while head is not None:
13             length += 1
14             head = head.next
15         return length
16             
17     def getIntersectionNode(self, headA, headB):
18         if headA is None or headB is None:
19             return None
20         
21         hA = headA
22         hB = headB
23         
24         lenA = self.getListLen(hA)
25         lenB = self.getListLen(hB)
26         
27         if lenA > lenB :
28             distance = lenA - lenB
29             for i in range(0,distance):
30                 hA = hA.next
31         if lenA < lenB :
32             distance = lenB -lenA
33             for i in range(0,distance):
34                 hB = hB.next
35         
36         intersection = None
37         while hA is not None and hB is not None:
38             if hA == hB:
39                 return hA
40             else:
41                 hA = hA.next
42                 hB = hB.next
43         return intersection

思路：

1. 首先要记录两个Linked List的长度

2. 双指针分别指向两个List 指向长List的先走若干步,获得一个新的Linked List表头

3. 逐一比较两个List的每个指针的值是否相等：直到找到相等的，或者到None

相关阅读:
JS・TextArea 字符串长度限制
JavaScript面试题目集锦
IE内存泄露分析
JS动态添加样式和脚本
取消锚(<a/>)点击后页面跳转的几种方法
JS代码片段整理
IE缓存问题的解决方法
字符串与数字转换
如何安装和配置Cassandra
字类和超类的转化问题

原文地址：https://www.cnblogs.com/xbf9xbf/p/4187862.html