leetcode之判断两链表首次交汇节点

题目：

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：先遍历左链表，再遍历右链表，遍历的时候分别计数，遍历完成后，如果最后一个节点不同，则没有交点，否则，可判断有交点，在有交点的情况下，看两条链表的便利次数M,N，然后再重新遍历，不过这次遍历先让长的那条链表先走|M-N|步，然后两边同时向前，每一步判断节点地址是否相等，第一次相等的即为首个交汇点。逻辑就是这样，但是这里有一些边界条件必须注意，否则，有了这个思路一样会出错。

代码

/** * 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* p1=headA;        ListNode* p2=headB;        if(p1==NULL||p2==NULL)        {            return NULL;        }        if(p1==p2)        {            return p1;        }        int i=1;        int j=1;        while(p1->next!=NULL)        {            p1=p1->next;            i++;        }        while(p2->next!=NULL)        {            p2=p2->next;            j++;        }        //no intersection, last node not the same        if(p1!=p2)        {            return NULL;        }        //else        p1=headA;        p2=headB;        if(i>j)        {            int diff=i-j;            while(diff>0)            {                p1=p1->next;                diff--;            }        }else if(j>i)        {            int diff=j-i;            while(diff>0)            {                p2=p2->next;                diff--;            }        }        if(p1==p2)        {            return p1;        }        while(p1->next!=p2->next)        {            p1=p1->next;            p2=p2->next;        }        return p1->next;    }};