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.

思路：假设A与B长度相等，则从链表头开始一一对比即可。

假设A与B长度不等，则进行一一对比时，短的链表会先被遍历完。这时，让该指针指向另一个链表的表头，继续进行。当长链表也被遍历完时，则其指针指向短的链表。通过这种方式，一定可以找到回合点，因为两个指针遍历的总长度是相等的，都是长链表和短链表各走了一遍。

 1 class Solution {
 2 public:
 3     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
 4         ListNode *ha = headA;
 5         ListNode *hb = headB;
 6         if (ha == NULL || hb == NULL) return NULL;
 7         while (ha != NULL && hb != NULL && ha != hb)
 8         {
 9             ha = ha->next;
10             hb = hb->next;
11             if (ha == hb) return ha;
12             if (ha == NULL) ha = headB;
13             if (hb == NULL) hb = headA;
14         }
15         return ha;
16     }
17 };