本文共 2020 字，大约阅读时间需要 6 分钟。

1 问题

判断链表是否包含环

2 思路

2个指针，一个指针走一步，一个指针走2步，如果相遇则有，反之无。

3 代码实现

#include 
   
    #include 
    
     #define true 1#define false 0;typedef struct node{    int value;    struct node *next;}Node;/* *判断链表是否有环 */int isCircleList(Node *head){    if (head == NULL)    {        return false;    }    Node *first = NULL;    Node *second = NULL;    first = head;    second = head;    while (second != NULL && (second->next) != NULL && (second->next->next != NULL))    {        first = first->next;        second = second->next->next;        if (first == second)        {            return true;        }    }    return false;}int main(){    Node *head = NULL;    Node *node1 = NULL;    Node *node2 = NULL;    Node *node3 = NULL;    Node *node4 = NULL;    Node *node5 = NULL;    Node *node6 = NULL;    Node *node7 = NULL;    head = (Node *)malloc(sizeof(Node));    node1 = (Node *)malloc(sizeof(Node));    node2 = (Node *)malloc(sizeof(Node));    node3 = (Node *)malloc(sizeof(Node));    node4 = (Node *)malloc(sizeof(Node));    node5 = (Node *)malloc(sizeof(Node));    node6 = (Node *)malloc(sizeof(Node));    node7 = (Node *)malloc(sizeof(Node));    if (head == NULL || node1 == NULL || node2 == NULL || node3 == NULL        || node4 == NULL || node5 == NULL || node6 == NULL || node7 == NULL)    {        printf("malloc fail\n");        return false;    }    //             node7<-node6 <-node5    //              |              |    //head->node1->node2->node3->node4    head->value = 0;    head->next = node1;    node1->value = 1;    node1->next = node2;    node2->value = 2;    node2->next = node3;    node3->value = 3;    node3->next = node4;    node4->value = 4;    node4->next = node5;    node5->value = 5;    node5->next = node6;    node6->value = 6;    node6->next = node7;    node7->value = 7;    node7->next = node2;    int result = isCircleList(head);    if (result)    {        printf("list have circle\n");    }    else    {        printf("list do not have circle\n");    }    return true;}
    
   

4 运行结果

list have circle

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