方法一：暴力法+求长度法(不推荐)
函数代码：
class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *p=headA; int len1=0; ListNode *q=headB; int len2=0; int dis=0; while(p) { p=p->next; len1++; } while(q) { q=q->next; len2++; } p=headA; q=headB; if(len1>len2) { dis=len1-len2; while(dis--) { p=p->next; } } else { dis=len2-len1; while(dis--) { q=q->next; } } while(p!=q) { p=p->next; q=q->next; } return p; } };方法二推荐：循环遍历+头尾相接法
class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *p=headA; ListNode *q=headB; while(p!=q) { p=p==NULL?headB:p->next; q=q==NULL?headA:q->next; } return p; } };