判断两个链表是否相交（编程之美3.6）

#include "stdafx.h"

struct Node
{
	int value;
	Node * next;
};

//判断是否有环，返回bool，如果有环，返回环里的节点
bool isCircle(Node * head, Node *& circleNode, Node *& lastNode)
{
	Node * fast = head->next;
	Node * slow = head;
	while(fast != slow && fast && slow)
	{
		if(fast->next != NULL)
			fast = fast->next;

		if(fast->next == NULL)
			lastNode = fast;
		if(slow->next == NULL)
			lastNode = slow;

		fast = fast->next;
		slow = slow->next;
		
	}
	if(fast == slow && fast && slow)
	{
		circleNode = fast;
		return true;
	}
	else
		return false;
}


bool detect(Node * head1, Node * head2)
{
	Node * circleNode1;
	Node * circleNode2;
	Node * lastNode1;
	Node * lastNode2;

	bool isCircle1 = isCircle(head1,circleNode1, lastNode1);
	bool isCircle2 = isCircle(head2,circleNode2, lastNode2);

	//一个有环，一个无环
	if(isCircle1 != isCircle2)
		return false;
	//两个都无环，判断最后一个节点是否相等
	else if(!isCircle1 && !isCircle2)
	{
		return lastNode1 == lastNode2;
	}
	//两个都有环，判断环里的节点是否能到达另一个链表环里的节点
	else
	{
		Node * temp = circleNode1;
		while(temp != circleNode1)
		{
			if(temp == circleNode2)
				return true;
			temp = temp->next;
		}
		return false;
	}

	return false;
}

int _tmain(int argc, _TCHAR* argv[])
{
	Node * n1 = new Node();
	Node * n2 = new Node();
	Node * n3 = new Node();
	Node * n4 = new Node();

	n1->next = n2;
	n2->next = n3;
	n3->next = n4;
	n4->next = NULL;


	Node * n5 = new Node();
	Node * n6 = new Node();
	Node * n7 = new Node();
	Node * n8 = new Node();

	n5->next = n6;
	n6->next = n7;
	n7->next = n8;
	n8->next = n5;

	if(detect(n1,n2))
		printf("相交\n");
	else
		printf("不相交\n");

	return 0;
}