【leetcode】Intersection of Two Linked Lists

 

Intersection of Two Linked Lists

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.


Notes:

  • If the two linked lists have no intersection at all, return null.
  • The linked lists must retain their original structure after the function returns.
  • You may assume there are no cycles anywhere in the entire linked structure.
  • Your code should preferably run in O(n) time and use only O(1) memory.
 
先从头到尾扫描,看两个链表长度相差多少。
然后让长的链表先走过这个长度差,然后两个链表同时走,直到相遇
 
 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
12        
13        
14         if(headA==NULL||headB==NULL)
15         {
16             return NULL;
17         }
18        
19         ListNode *hA=headA,*hB=headB;
20        
21         int lenA=1,lenB=1;
22         while(hA->next!=NULL)
23         {
24             lenA++;
25             hA=hA->next;
26         }
27        
28         while(hB->next!=NULL)
29         {
30             lenB++;
31             hB=hB->next;
32         }
33        
34        
35         if(hA!=hB)
36         {
37             return NULL;
38         }
39        
40         int dis=lenA-lenB;
41        
42        
43         hA=headA;
44         hB=headB;
45        
46         if(dis>0)
47         {
48             while(dis)
49             {
50                 hA=hA->next;
51                 dis--;
52             }
53         }
54        
55         if(dis<0)
56         {
57             dis=-dis;
58             while(dis)
59             {
60                 hB=hB->next;
61                 dis--;
62             }
63         }         
64        
65         while(hA!=hB)
66         {
67             hA=hA->next;
68             hB=hB->next;
69         }
70        
71        
72         return hA;
73     }
74 };
75  

 

posted @ 2015-01-26 23:29  H5开发技术  阅读(233)  评论(0编辑  收藏  举报