leetcode 160

160. 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.

找到两个单链表相交的点。此处假设单链表不存在环。

思路：首先获得两个单链表的长度，得到长度的差值n，然后将指向长链表的指针A先向后移动n位，之后指针A同指向短链表的指针B同时每次移动一位，当A与B指向同一节点时，该节点就是相交的结点。

 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         if(headA == NULL || headB == NULL)
13         {
14             return NULL;
15         }
16         ListNode * a = headA;
17         ListNode * b = headB;
18         int n = 1;
19         int m = 1;
20         while(headA->next != NULL)
21         {
22             headA = headA->next;
23             n++;
24         }
25         while(headB->next != NULL)
26         {
27             headB = headB->next;
28             m++;
29         }
30         if(headA != headB)
31         {
32             return NULL;
33         }
34         if(n > m)
35         {
36             for(int i = 0; i < n-m; i++)
37             {
38                 a = a->next;
39             }
40             while(a)
41             {
42                 if(a == b)
43                 {
44                     return a;
45                 }
46                 else
47                 {
48                     a = a->next;
49                     b = b->next;
50                 }
51             }
52         }
53         else
54         {
55             for(int i = 0; i < m-n; i++)
56             {
57                 b = b->next;
58             }
59             while(b)
60             {
61                 if(a == b)
62                 {
63                     return a;
64                 }
65                 else
66                 {
67                     a = a->next;
68                     b = b->next;
69                 }
70             }
71         }
72         return NULL;
73     }
74 };

拓展：

http://www.cppblog.com/humanchao/archive/2015/03/22/47357.html