剑指offer之判断链表是否包含环

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 <stdio.h>
#include <stdlib.h>#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->node4head->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;
}