算法导论 10-2-8 个指针实现双链表

最新推荐文章于 2024-04-20 21:32:56 发布

原创最新推荐文章于 2024-04-20 21:32:56 发布 · 887 阅读

2 ·

CC 4.0 BY-SA版权

算法导论专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种使用单个指针字段实现双向链表的方法，并提供了完整的C++代码实现，包括插入、查找和删除操作。通过XOR运算巧妙地在每个节点中存储前后节点的地址。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：利用一个指针实现双链表。每一个节点里面存一个np，x.np = x.pre XOR x.next

思路： a XOR a = 0； 0 XOR a = a; 因此，np与父节点进行xor运算之后，就能得到next的地址。

#include <iostream>
using namespace std;
struct LinkList{
    int key;
    int prenext;
};

struct LinkList* insertLinkList(struct LinkList*head,int d,struct LinkList**tail);
struct LinkList* searchLinkList(struct LinkList*head,int d);
struct LinkList* deleteLinkList(struct LinkList*head,int d);
void print(struct LinkList* head);
int main ()
{
    struct LinkList*head = 0;
    struct LinkList*tail = 0;
    for (int i = 0;i < 12;i ++)
    {
        head = insertLinkList(head,i,&tail);
        //cout << tail->key;
    }
    print(head);
    print(tail);
    struct LinkList* tr = searchLinkList(head,12);
    if (tr)
    {
        cout << tr-> key << endl;
    }
}
struct LinkList* insertLinkList(struct LinkList*head,int d,struct LinkList**tail)
{
    struct LinkList * tmp = new struct LinkList();
    tmp->key = d;
    tmp->prenext = 0;
    if (NULL == head)
    {
        head = tmp;
        head->prenext = 0;
        *tail = tmp;
        return head;
    }
    if ( 0 == head->prenext)
    {
        head->prenext = (int)tmp;
        tmp->prenext = tmp->prenext^((int)head);
        *tail = tmp;
        return head;
    }
    struct LinkList * p = head;
    struct LinkList * q = (struct LinkList *)head->prenext;
    struct LinkList * r = NULL;
    struct LinkList * s = NULL;
    while (q != 0)
    {
        r = q;
        s = p;
        q = (struct LinkList*)(q->prenext^((int)p));
        p = r;
        r = s;
    }
    p->prenext = ((int)tmp)^((int)r);
    tmp->prenext = ((int)p);
    *tail = tmp;
    return head;

}
void print(struct LinkList* head)
{
    if (NULL == head)
    {
        return ;
    }
    struct LinkList * p = head;
    cout << p->key << ' ';
    struct LinkList * q = (struct LinkList *)head->prenext;
    if (NULL == q)
    {
        return;
    }
    struct LinkList * r = NULL;
    struct LinkList * s = NULL;
    while (q != NULL)
    {
        r = (struct LinkList *)(q->prenext^((int)p));
        cout << q->key << ' ';
        p = q;
        q = r;
    }
    cout << endl;
}
struct LinkList* searchLinkList(struct LinkList*head,int d)
{
    if (NULL == head)
    {
        return 0;
    }
    struct LinkList * p = head;
    if (p->key == d)
    {
        return p;
    }
   // cout << p->key << ' ';
    struct LinkList * q = (struct LinkList *)head->prenext;
    if (NULL == q)
    {
        return 0;
    }
    struct LinkList * r = NULL;
    struct LinkList * s = NULL;
    while (q != NULL)
    {
        r = (struct LinkList *)(q->prenext^((int)p));
       if (q->key == d)
       {
           s = q;
           break;
       }
        //cout << q->key << ' ';
        p = q;
        q = r;
    }
    return s;
  //  cout << endl;
}
struct LinkList* deleteLinkList(struct LinkList*head,int d)
{

}