Algorithm One Day One -- 判斷鏈表是否有環(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)
{
	//建立一個鏈表	
	Node *head,*tail,*cnew;	
	head=NULL;	
	int num;	
	cout <<"輸入數據(以#鍵結束):" << endl;
	while(1 )
	{
	cin >>num ;	
	if('#'==getchar())
	//以#鍵表示輸入結束	  
	break;
	cnew=new  Node;;	
	cnew->data=num;	 
	cnew->next=NULL;	
	if(head==NULL)
		//若爲空則將頭節點指向新節點	 
		head=cnew;
	else	 
	   tail->next=cnew;
	//將當前節點的next指向新的節點	
	tail=cnew;	
   (*numNode)++;	
	}
	return head;}

/*判斷是否有環思路概述：分別定義步長爲1和2的指針fast and slow
指向頭結點，if無環，則fast先走到終點；若是鏈表長度爲奇數時，
fast->Next爲空；當鏈表長度爲偶數時，fast爲空*/

bool isLoop(pNode  pHead)
{
	pNode fast = pHead;
	pNode slow = pHead;
	while( fast != NULL && fast->next != NULL)
	{
		fast = fast->next->next;
		slow = slow->next;
		//若是有環，則fast會超過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;
	//初始化將節點個數初始化爲零
   pNode head=NULL ;
   	cout <<"鏈表head的節點個數爲: "  <<endl;
   cin >>numnode;
	head=Create(&numnode);	
    isLoop(head);
	return 0;
}