380. Intersection of Two Linked Lists【medium】

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

Notice

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.

Example

The following two linked lists:

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

begin to intersect at node c1.

Challenge

Your code should preferably run in O(n) time and use only O(1) memory.

解法一：

 1 class Solution {
 2 public:
 3     /**
 4      * @param headA: the first list
 5      * @param headB: the second list
 6      * @return: a ListNode
 7      */
 8     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
 9         // write your code here
10         if(headA == NULL || headB == NULL)
11             return NULL;
12         ListNode* iter1 = headA;
13         ListNode* iter2 = headB;
14         int len1 = 1;
15         while(iter1->next != NULL)
16         {
17             iter1 = iter1->next;
18             len1 ++;
19         }
20         int len2 = 1;
21         while(iter2->next != NULL)
22         {
23             iter2 = iter2->next;
24             len2 ++;
25         }
26         if(iter1 != iter2)
27             return NULL;
28         if(len1 > len2)
29         {
30             for(int i = 0; i < len1-len2; i ++)
31                 headA = headA->next;
32         }
33         else if(len2 > len1)
34         {
35             for(int i = 0; i < len2-len1; i ++)
36                 headB = headB->next;
37         }
38         while(headA != headB)
39         {
40             headA = headA->next;
41             headB = headB->next;
42         }
43         return headA;
44     }
45 };