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    }};