leetcode 【 Intersection of Two Linked Lists 】python 实现

本文提供了一种寻找两个单链表开始相交节点的算法实现。通过计算两个链表的长度,并利用双指针技巧来定位交点,最终返回交点的节点。此算法已通过在线测试验证。

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

转载于:https://www.cnblogs.com/xbf9xbf/p/4187862.html

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