51-Intersection of Two Linked Lists

1. Intersection of Two Linked Lists My Submissions QuestionEditorial Solution
   Total Accepted: 72580 Total Submissions: 239762 Difficulty: Easy
   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.

思路：两条链表如果相交，那从交点开始，后面一定都是相等的
那如何求第一个交点呢？
时间复杂度：O(n)
要么返回空
有交点时，说明尾部是对齐的，要找到第一个交点，只要让长的链表先走len1-len2步，这里假设len1是那条长链，那么此时再同时走，相遇的第一个节点便是交点

/** * 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;//有一为空返回
        if(headA==headB)return headA;//头部相同返回
        ListNode *l1=headA,*l2=headB;
        int len1=1,len2=1;
        while(l1->next){//遍历记录长度
            l1=l1->next;
            len1++;
        }
        while(l2->next){
            l2=l2->next;
            len2++;
        }
        ListNode *p = len1>len2?headA:headB;    //p为长链表
        ListNode *psmall=len1>len2?headB:headA;    //psmall为短链表
        int count=abs(len2-len1);
        while(count--){
                p=p->next;
        }
        while(p!=NULL&&psmall!=p){
            p=p->next;
            psmall = psmall->next;
        }
        if(p!=NULL)return p;          //有交点
        else return NULL;             //无交点

    }
};