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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

1）把其中的一条链首尾相连，组成一个环，这样问题就转化成：判断一条链表是否有环，如果有，求环的入口问题。

2）把两个链表都首尾颠倒顺序，然后比较。比较完之后，再颠倒把链表的顺序，恢复原样。

算法1的代码：

c++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(!headA||!headB)
            return NULL;
        ListNode *p = headB;
        while(p->next){
            p=p->next;
        }
        ListNode *tailB = p;
        tailB->next=headB;
        
        ListNode *p1,*p2;
        p1=headA;
        p2=headA;
        while(p1){
            p1=p1->next;
            p2=p2->next;
            if(p1){
                p1=p1->next;
            }else{
                tailB->next = NULL;
                return NULL;
            }
            if(p1==p2){
                break;
            }
        }
        if(p1==NULL){
            tailB->next = NULL;
            return NULL;
        }
        p2=headA;
        while(p1!=p2){
            p2=p2->next;
            p1=p1->next;
        }
        tailB->next = NULL;
        return p2;
        
    }
};

Intersection of Two Linked Lists