本来想自己写，写了一半发现一篇文章，解释写得简单易懂，我就直接拿过来了。

这个问题值得反复地写，锻炼链表coding能力的好题。

 

 

//如果两个链表都不带环
int NotCycleCheckCross(pLinkNode head1,pLinkNode head2)
{pLinkNode list1 = head1->next;pLinkNode list2 = head2->next;if ((NULL==list1 )||(NULL==list2)){return 0;      //不相交}while (NULL != list1->next){list1 = list1->next;}while (NULL != list2->next){list2 = list2->next;}if (list1==list2){return 1;      //相交}return 0;          //不相交
}//链表带环，判断两个链表是否相交
int CycleCheckCross(pLinkNode meet1, pLinkNode meet2)
{pLinkNode cur = meet1->next;if (meet1 == meet2){return 1;     //链表相交}while ((cur != meet1)&&(cur!=meet2)){cur = cur->next;}if (cur == meet2){return 1;   //链表相交}return 0;       //不相交
}
//将上面两个函数封装成一个函数int CheckCross(pLinkNode head1, pLinkNode head2)          //参数为两个头结点{pLinkNode fast = NULL;pLinkNode slow = NULL;pLinkNode meet1 = NULL;pLinkNode meet2 = NULL;if (head1->next == NULL || head2->next == NULL){return 0;          //至少一个链表为空链表，则两个链表一定不相交}fast = head1->next;slow = head1->next;while (fast&&fast->next)           //判断链表head1是否带环{fast = fast->next->next;slow = slow->next;if (fast == slow){meet1 = fast;break;}}fast = head2->next;slow = head2->next;while (fast&&fast->next)           //判断链表head2是否带环{fast = fast->next->next;slow = slow->next;if (fast == slow){meet2 = fast;break;}}if ((meet1 == NULL) && (meet2 == NULL))       //如果两个链表都不带环{return NotCycleCheckCross(head1, head2);}else if (meet1&&meet2)                        //如果两个链表都带环{return CycleCheckCross(meet1, meet2);}//如果两个链表一个带环一个不带环，则一定不相交直接返回0return 0;            //不相交}

原文https://blog.csdn.net/lf_2016/article/details/51756644