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.\n");
else
printf("method1: has no loop.\n");

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

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