链表的交叉节点：不包含环

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.

拓展思维：链表调换顺序可以考虑使用“栈”来操作、

/**
 * 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) {
        ListNode*pA=headA;
        ListNode*pB=headB;
        int lenA=0,lenB=0;
        while(pA){
            ++lenA;
            pA=pA->next;
        }
        while(pB){
            ++lenB;
            pB=pB->next;
        }
        pA=headA;pB=headB;
        
        if(lenAnext;
        }else {
            int k=0;
            while(k++next;
        }

        while(pA!=pB){
            pA=pA->next;
            pB=pB->next;
        }
        return pA;
    }
};