160. Intersection of Two Linked Lists leetcode （list）

1. 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.

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {        if(NULL == headA || NULL == headB) //如果两个链表有一个为空则返回空，不存在相交的点        {            return NULL;        }        ListNode *p1 = headA;        ListNode *p2 = headB;        ListNode * result=NULL; //用来保存结果的节点        int len1 = 0;        int len2 = 0;        while(p1) //遍历获得链表1的长度        {            len1++;            p1 = p1->next;        }        while(p2)//遍历获得链表2的长度        {            len2++;            p2 = p2->next;        }        if(len2 > len1)//使得p1始终指向长的链表，p2指向短的链表        {            p1 = headB;            p2 = headA;        }        else        {            p1 = headA;            p2 = headB;        }        int len = abs(len1 - len2);        while(len) //长的链表先走完长的节点        {            p1 = p1->next;            len--;        }        while(p1&&p2)        {            if(p1->val == p2->val) //如果相等则找到相交的节点            {                result = p1;                break;            }            else            {                p1 = p1->next;                p2 = p2->next;            }        }        return result;    }

//最开始提交的代码错误的原因，我去~~~~~~~ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {        if(NULL == headA || NULL == headB)        {            return NULL;        }        ListNode *p1 = headA;        ListNode *p2 = headB;        int len1 = 0;        int len2 = 0;        while(p1)        {            len1++;            p1 = p1->next;        }        while(p2)        {            len2++;            p2 = p2->next;        }        p1 = headA; //注意这两句不能少，开始就是因为缺少这两句，我去~~~~~~~~~~~        p2 = headB;        int num = abs(len1 - len2);        while(num)        {            if(len1 >len2)            {                p1 = p1->next;                //cout<<" 1 p1->val="<<p1->val<<" ";            }            else            {                p2 = p2->next;            }            num--;        }        while(p1 && p2)        {            if(p1->val== p2->val)            {                return p1;                break;            }            p1 = p1->next;            p2 = p2->next;        }        return NULL;    }