链表相交

判断两个链表是否相交？？？

我们先来分析一下两个链表的相交问题，如图：

两条链表都不带环：

一条链表带环，一条不带环：

两条链表都带环：

//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0
	return 0;            //不相交
}