Question

我在创建循环双向链表时遇到问题。添加新节点时，我的代码似乎工作正常。它在删除每个其他节点时也有效。当我尝试擦除2个连续节点时出现问题。在过去的两天里，我已经查看了我的代码，虽然我发现了一些错误但没有更正问题。

void LinkedList::insert(Node *newnode, Node * pred)
{
    Node *after;

    if ((empty()) && (pred != head))
    {
        cout << "*&$MU##\n";
    }
    else
    {
        after = pred->next;

        newnode->next = pred->next;
        newnode->pred = pred;
        pred->next->pred = newnode;
        pred->next = newnode;

        listSize++;
    }
}

void LinkedList::erase(Node *pred)
{   
    Node *actual; //actual node to be deleted
    Node *after; //node after the node to be deleted
    if (!empty())
    {
        actual = pred->next;

        after = actual->next;
        pred->next = after;
        after->pred = pred;
        actual->next = '\0';
        actual->pred = '\0';

        listSize--;
    }
}

添加节点的命令如下：mylist.insert（newnode，pred）擦除节点的命令是：erase（pred）

我相信当我擦除节点时会出现问题，因为我的函数将pred和next值设置为NULL。当我使用以下节点再次调用擦除功能时，它不能再执行以下actual = pred->next; after = actual->next;，因为前一个命令已将该节点的pred和next设置为NULL。

任何帮助将不胜感激。

我已决定为我的LinkedList.cpp添加整个代码，以防出现问题。

#include "stdafx.h"
#include <iostream>
using namespace std;
#include "Node.h"
#include "LinkedList.h"

LinkedList::LinkedList()
{
    Node * hnode = new Node('?');
    head = hnode;
    listSize = 0;
    first = head;
    first->next = first;
    //first->pred = first;
}
bool LinkedList::empty()//O(1)
{
    if (first->next == first)
        return 1;
    else
        return 0;
}

void LinkedList::insert(Node *newnode, Node * pred)
{
    Node *after;

    if ((empty()) && (pred != head))
    {
        cout << "*&$MU##\n";
    }
    else
    {
        after = pred->next;

        newnode->next = pred->next;
        newnode->pred = pred;
        pred->next->pred = newnode;
        pred->next = newnode;

        listSize++;
    }
}
void LinkedList::erase(Node *pred)
{   
    Node *actual; //actual node to be deleted
    Node *after; //node after the node to be deleted
    if (!empty())
    {
        actual = pred->next;
        after = actual->next;

        pred->next = after;
        after->pred = pred;
        actual->next = '\0';
        actual->pred = '\0';

        listSize--;
    }
}

void LinkedList::display()
{
    Node * current_node;
    current_node = first;
    do
    {
        cout << current_node->data << " ";
        current_node = current_node->next;
    } while (current_node != first);
    cout << listSize<< '\n';
}

在Doubly Linked list C ++中插入和删除节点

0 个答案: