Algorithm One Day One -- 判断链表是否有环(上)



Is a loop ? Question descrip as follows :

Assume that wehave a head pointer to a link-list. Also assumethat we know the list is single-linked. Can you come up an algorithm to checkwhether this link list includes a loop by using O(n) time and O(1) space wheren
is the length of the list? Furthermore, can you do so with O(n) time and onlyone register?

/********************************************************************
created:2015年1月22日 00:54:56
author: Jackery
purpose: Is there a loop ?
*********************************************************************/
#include"stdafx.h"
#include<iostream>
using namespace std;

typedef struct node
{
	int data;
	 struct node *next;
}Node,*pNode;

Node *Create(int *numNode)
{
	//创建一个链表
	Node *head,*tail,*cnew;
	head=NULL;
	int num;
	cout <<"输入数据(以#键结束):" << endl;
	while(1 )
	{
	cin >>num ;
	if('#'==getchar())
	//以#键表示输入结束
	break;
	cnew=new  Node;;
	cnew->data=num;
	cnew->next=NULL;
	if(head==NULL)
		//若为空则将头节点指向新节点
		head=cnew;
	else
	   tail->next=cnew;
	//将当前节点的next指向新的节点
	tail=cnew;
   (*numNode)++;
	}
	return head;}

/*判断是否有环思路概述：分别定义步长为1和2的指针fast and slow
指向头结点，if无环，则fast先走到终点；如果链表长度为奇数时，
fast->Next为空；当链表长度为偶数时，fast为空*/

bool isLoop(pNode  pHead)
{
	pNode fast = pHead;
	pNode slow = pHead;
	while( fast != NULL && fast->next != NULL)
	{
		fast = fast->next->next;
		slow = slow->next;
		//如果有环，则fast会超过slow一圈
		if(fast == slow)
		{
			break;
		}
	}

	if(fast == NULL || fast->next == NULL  )
	{
		cout <<"Wow,there is not loop in the list "<< endl;
		return false;
	}
	else
	{
		cout <<"Yeah,there is loop in the list " << endl;
		return true;
	}
}
int main(int argc ,char * argv[])
{
	int numnode=0;
	//初始化将节点个数初始化为零
   pNode head=NULL ;
   	cout <<"链表head的节点个数为: "  <<endl;
   cin >>numnode;
	head=Create(&numnode);
    isLoop(head);
	return 0;
}