判断单链表是否有环

来源：互联网发布：app编程培训编辑：程序博客网时间：2024/05/22 14:49

算法思想: A跑一步,B跑两步, 如果有环形区, B一定会与A相遇.

 #include <iostream> #include <string> using namespace std;  typedef struct list_node {     int x;     struct list_node* next; }node, *pnode;bool isCircle(pnode list){    pnode quick=list, slow=list;//快慢指针    while(1)    {        if(quick==NULL || slow == NULL)//可以到达链表末尾说明无环            return false;        quick = quick->next;        if(quick == NULL)            return false;        quick = quick->next;        slow = slow->next;        if(slow==quick)//慢指针赶上快指针说明有环            return true;    }}int main(){    pnode list = NULL, temp;    for(int i=1;i<=10;i++)    {        pnode pn = new node;        pn->x = i;        pn->next = list;        list = pn;    }        temp = list;    for(int i=0;i<10;i++, temp=temp->next)        cout<<temp->x<<" ";    cout<<endl;    if(isCircle(list))        cout<<"有环"<<endl;    else        cout<<"无环"<<endl;    //手工构造一个环    pnode tail = list;    while(tail->next != NULL)        tail = tail->next;    tail->next = list;    if(isCircle(list))        cout<<"有环"<<endl;    else        cout<<"无环"<<endl;    for(int i=0;i<10;i++)    {        temp = list->next;        delete list;        list = temp;    }    return 0;}