• 数据结构-判断链表是否存在环形链表


    1:这里有一个比较简单的解法。设置两个指针p1、p2。每次循环p1向前走两步。直到p2碰到NULL指针或者两个指针相等时结束循环。如果两个指针相等,则说明存在环。代码如下:

    // ConsoleApplication15.cpp : 定义控制台应用程序的入口点。
    //
    
    #include "stdafx.h"
    #include <malloc.h>
    #include <iostream>
    using namespace std;
    
    typedef struct node//定义链表结构体
    {
        int data;//节点内容
        node *next;//指向结构体的指针,下一个节点
    }node;
    
    node *create()//创建单链表
    {
        int i = 0;//链表中数据的个数
        node *head, *p, *q;//这些的本质是节点的地址
        int x = 0;
        head = NULL;
        q = NULL;//初始化q,q代表末节点
        p = NULL;
        while (1)
        {
            printf("please input the data:");
            scanf_s("%d", &x);
            if (x == 0)
                break;//data为0时创建结束
            p = (node *)malloc(sizeof(node));//用于每次输入链表的数据
            p->data = x;
            if (++i == 1)//链表头的指针指向下一个节点
            {
                head = p;
                q = p;
            }
            else
            {
                q->next = p;//连接到链表尾端
                q = p;
            }
            q->next = NULL;/*尾结点的后继指针为NULL(空)*/
        }
        return head;
    }
    
    int length(node *head)
    {
        int len = 0;
        node *p;
        p = head->next;
        while (p != NULL)
        {
            len++;
            p = p->next;
        }
        return len;
    }
    
    void print(node *head)
    {
        node *p;
        p = head;
        while (p)/*直到结点q为NULL结束循环*/
        {
            printf("%d ", p->data);/*输出结点中的值*/
            p = p->next;/*指向下一个结点*/
        }
    }
    
    node *search_node(node *head, int pos)//查找单链表pos位置的节点,返回节点的指针。pos从0开始,0返回head节点
    {
        node *p = head->next;
        if (pos < 0)//pos位置不正确
        {
            printf("incorrect position to search node!");//pose位置不正确
            return NULL;
        }
        if (pos == 0)
        {
            return head;
        }
        if (pos == NULL)
        {
            printf("Link is empty!");//链表为空
            return NULL;
        }
        while (--pos)
        {
            if ((p = p->next) == NULL)
            {
                printf("incorrect position to search node!");//超出链表返回
                break;
            }
        }
        return p;
    }
    
    node *insert_node(node *head, int pos, int data)//单链表的插入
    {
        node *item = NULL;
        node *p;
        item = (node *)malloc(sizeof(node));
        item->data = data;
        if (pos == 0)//插在head后面
        {
            head->next = item;//head后面是item
            return head;
        }
        p = search_node(head, pos);//获得pos的节点指针
        if (p != NULL)
        {
            item->next = p->next;//item指向原pos节点的后一个节点
            p->next = item;//把item插入到pos的后面
        }
        return head;
    }
    
    node *delete_node(node *head, int pos)//删除节点
    {
        node *item = NULL;
        node *p = head->next;
        if (p = NULL)
        {
            printf("link is empty!");
            return NULL;
        }
        p = search_node(head, pos - 1);//获得位置pos节点的指针
        if (p != NULL&&p->next != NULL)
        {
            item = p->next;
            p->next = item->next;
            delete item;
        }
        return head;
    }
    
    node *reverse(node *head)//链表的逆置
    {
        node *next;
        node *prev = NULL;
        while (head != NULL)
        {
            next = head->next;
            head->next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
    
    node *search(node *head)//寻找单链表的中间元素
    {
        int i = 0;
        int j = 0;
        node *current = NULL;
        node *middle = NULL;
        current = middle = head->next;
        while (current != NULL)
        {
            if (i / 2 > j)
            {
                j++;
                middle = middle->next;
            }
            i++;
            current = current->next;
        }
        return middle;
    }
    
    node *InsertSort(void)//单链表的正向排序
    {
        int data = 0;
        struct node *head = NULL, *New, *Cur, *Pre=NULL;
        while (1)
        {
            printf("please input the data
    ");
            scanf_s("%d", &data);
            if (data == 0)
            {
                break;
            }
            New = (struct node*)malloc(sizeof(struct node));
            New->data = data;
            New->next = NULL;
            if (head == NULL)//第一次循环时对头节点赋值
            {
                head = New;
                continue;
            }
            if (New->data <= head->data)
            {//head之前插入节点
                New->next = head;
                head = New;
                continue;
            }
            Cur = head;
            while (New->data > Cur->data && Cur->next != NULL)//找到需要插入的位置
            {
                Pre = Cur;
                Cur = Cur->next;
            }
            if (Cur->data >= New->data)//位置在中间
            {
                Pre->next = New;
                New->next = Cur;
            }
            else//位置在末尾
                Cur->next = New;//把New节点插到Cur之后
        }
        return head;
    }
    
    bool IsLoop(node *head, node **start)//判断链表是否存在环形链表,start为回环开始节点的地址
    {
        node *p1 = head, *p2 = head;
        if (head == NULL || head->next == NULL)//head为NULL或者链表为空时返回false
        {
            return false;
        }
        do
        {
            p1 = p1->next;//p1走一步
            p2 = p2->next->next;//p2走两步
        } while (p2 && p2->next && p1 != p2);
        if (p1 == p2)
        {
            *start = p1;//p1为回环开始节点
            return true;
        }
        else
        {
            return false;
        }
    }
    
    int main()
    {
        bool bLoop = false;
        node *head = create();
        node *start = head->next->next->next;//使第4个节点为回环开始位置
        start->next = head->next;//回环连接到第2个节点
        node *loopStart = NULL;
        bLoop = IsLoop(head, &loopStart);
        printf("bLoop=%d
    ", bLoop);
        printf("bLoop==loopStart?%d
    ", (loopStart == start));
        return 0;
    
    
        
    
        return 0;
    }
    View Code

    运行结果:

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  • 原文地址:https://www.cnblogs.com/lovemi93/p/7600233.html
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