判断单链表是否有环的两种方法（转）

如图，如果单链表有环，则在遍历时，在通过6之后，会重新回到3，那么我们可以在遍历时使用两个指针，看两个指针是否相等。

方法一：使用p、q两个指针，p总是向前走，但q每次都从头开始走，对于每个节点，看p走的步数是否和q一样。如图，当p从6走到3时，用了6步，此时若q从head出发，则只需两步就到3，因而步数不等，出现矛盾，存在环
方法二：使用p、q两个指针，p每次向前走一步，q每次向前走两步，若在某个时候p == q，则存在环。

#include <stdio.h>
#include <stdlib.h>

#define LEN 8
typedef struct node* node_t;

struct node{  
    char val;  
    struct node *next;  
};  

//method 1
int has_loop(struct node *head);
//method 2
int has_loop2(node_t head);

int main()
{
    node_t* arr = (node_t*)malloc(sizeof(struct node)*LEN);
    arr[0] = (node_t)malloc(sizeof(struct node));
    int i;
    for(i = 1; i < LEN; i++)
    {
        arr[i] = (node_t)malloc(sizeof(struct node));
        arr[i - 1]->next = arr[i];
    }
    arr[LEN - 1]->next = NULL;

    //you can add a loop here to test
    //arr[6]->next = arr[0];
    if (has_loop(arr[0]))
        printf("method1: has loop.
");
    else
        printf("method1: has no loop.
");

    if (has_loop2(arr[0]))
        printf("method2: has loop.
");
    else
        printf("method2: has no loop.
");

    return 0;
}

//if two pointer are equal, but they don't have the same steps, then has a loop
int has_loop(node_t head)  
{  
    node_t cur1 = head;  
    int pos1 = 0;  
    while(cur1){  
        node_t cur2 = head;  
        int pos2 = 0;  
        pos1 ++;  
        while(cur2){  
            pos2 ++;  
            if(cur2 == cur1){  
                if(pos1 == pos2)  
                    break;  
                else  
                    return 1;
            }  
            cur2 = cur2->next;  
        }  
        cur1 = cur1->next;  
    }  
    return 0;  
} 

//using step1 and step2 here 
//if exists a loop, then the pointer which use step2 will catch up with the pointer which uses step1
int has_loop2(node_t head)
{
    node_t p = head;
    node_t q = head;
    while (p != NULL && q != NULL)
    {
        /*
        p = p->next;
        if (q->next != NULL)
            q = q->next->next;
        if (p == q)
            return 1;
            */
        //correct it on 17/11/2012
        p = p->next;
        q = q->next;
        if (q != NULL)
            q = q->next;
        if (p != NULL && p == q)
            return 1;
    }
    return 0;
}

原文转自 http://www.cnblogs.com/shuaiwhu/archive/2012/05/03/2480509.html