Intersection of Two Linked Lists——经典问题

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.

开始想道用栈来做，先入栈，然后出栈比较直接就出结果了，但是题目说只能利用一个空间，所以栈的方法不行了。

看网上的答案，没想到这么简单，暴力解法，什么算法、数据结构都没用，题刷多了，脑袋都僵了，一直以为要用什么算法来做呢。。。。

思路：

查找两个链表的第一个公共节点，如果两个节点的尾节点相同，肯定存在公共节点

方法： 长的链表开始多走 （h1的数量 - h2的数量）步，然后和短链表同步往下走，遇到的第一个相同的节点就是最早的公共节点

/**
 * 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==NULL||headB==NULL)
           return NULL;
        ListNode *flagA=headA;
        ListNode *flagB=headB;
        int countA=;
        int countB=;
        while(flagA->next!=NULL)
        {
            countA++;
            flagA=flagA->next;
        }
        while(flagB->next!=NULL)
        {
            countB++;
            flagB=flagB->next;
        }
        if(flagA!=flagB)
            return NULL;
        else
        {
            int diff=abs(countA- countB);
            if(countA>=countB)
            {
                flagA=headA;
                flagB=headB;
            }
            else
            {
                flagA=headB;
                flagB=headA;
            }
            while(diff)
            {
                flagA=flagA->next;
                diff--;
            }
            while(flagA!=flagB)
            {
                flagA=flagA->next;
                flagB=flagB->next;
            }
            return flagA;
        }
    }
};