判断单链表是否有环的两种方法

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。

代码如下：

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 
 4 #define LEN 8
 5 typedef struct node* node_t;
 6 
 7 struct node{  
 8     char val;  
 9     struct node *next;  
10 };  
11 
12 //method 1
13 int has_loop(struct node *head);
14 //method 2
15 int has_loop2(node_t head);
16 
17 int main()
18 {
19     node_t* arr = (node_t*)malloc(sizeof(struct node)*LEN);
20     arr[0] = (node_t)malloc(sizeof(struct node));
21     int i;
22     for(i = 1; i < LEN; i++)
23     {
24         arr[i] = (node_t)malloc(sizeof(struct node));
25         arr[i - 1]->next = arr[i];
26     }
27     arr[LEN - 1]->next = NULL;
28 
29     //you can add a loop here to test
30     //arr[6]->next = arr[0];
31     if (has_loop(arr[0]))
32         printf("method1: has loop.
");
33     else
34         printf("method1: has no loop.
");
35 
36     if (has_loop2(arr[0]))
37         printf("method2: has loop.
");
38     else
39         printf("method2: has no loop.
");
40 
41     return 0;
42 }
43 
44 //if two pointer are equal, but they don't have the same steps, then has a loop
45 int has_loop(node_t head)  
46 {  
47     node_t cur1 = head;  
48     int pos1 = 0;  
49     while(cur1){  
50         node_t cur2 = head;  
51         int pos2 = 0;  
52         pos1 ++;  
53         while(cur2){  
54             pos2 ++;  
55             if(cur2 == cur1){  
56                 if(pos1 == pos2)  
57                     break;  
58                 else  
59                     return 1;
60             }  
61             cur2 = cur2->next;  
62         }  
63         cur1 = cur1->next;  
64     }  
65     return 0;  
66 } 
67 
68 //using step1 and step2 here 
69 //if exists a loop, then the pointer which use step2 will catch up with the pointer which uses step1
70 int has_loop2(node_t head)
71 {
72     node_t p = head;
73     node_t q = head;
74     while (p != NULL && q != NULL)
75     {
76         /*
77         p = p->next;
78         if (q->next != NULL)
79             q = q->next->next;
80         if (p == q)
81             return 1;
82             */
83         //correct it on 17/11/2012
84         p = p->next;
85         q = q->next;
86         if (q != NULL)
87             q = q->next;
88         if (p != NULL && p == q)
89             return 1;
90     }
91     return 0;
92 }