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->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


 

作者：chen.yu
深信服三年半工作经验，目前就职游戏厂商，希望能和大家交流和学习,
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