Algorithm One Day One -- 判断链表是否有环(上)
来源：程序员人生 发布时间：2015-01-28 08:45:51 阅读次数：3019次
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1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;}
/*判断是不是有环思路概述：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slow1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;
}
生活不易，码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
