头未在链接列表中正确链接 [英] Head not linked correctly in linked list

问题描述

从链接列表中删除元素后显示时，将显示0以代替删除的元素。我有麻烦更新节点。任何人请解释发生了什么？为什么显示0？

When i display after removing a element from linked list, 0 is displayed in place of removed element. I am having trouble updating the nodes.Can anyone Please explain what is happening? Why 0 is displayed?

#include<iostream>
#include<stdlib.h>
using namespace std;

    class node {

    public:
      int data;
      node *link;

    };

    class linkedlist {

        node *head;

    public:

        linkedlist() {
            head=NULL;
        }

        int add(int data1) {

            node *insertnode=new node;
            insertnode->data=data1;
            insertnode->link=NULL;

            node *temp=head;

            if(temp!=NULL)
            {
                while(temp->link!=NULL)
                {
                    temp=temp->link;
                }
                temp->link=insertnode;

            }
            else{head=insertnode;}

        }

        void disp()
        {
            node *temp1=head;
            cout<<endl;
            if(temp1==NULL)
            {
                cout<<"Empty"<<endl;
            }

            if(temp1->link==NULL)
            {
                cout<<temp1->data<<endl;

            }
            else {

                do {
                    cout<<temp1->data<<endl;
                    temp1=temp1->link;
                } while(temp1!=NULL);

            }

        }

        int remove(int removedata)
        {
            node *previous;
            node *temp2=head;
            if(temp2==NULL)
                {exit(0);}

            if(temp2->link==NULL)
            {
                delete temp2;
                head=NULL;
            }

            else
            {

                while(temp2!=NULL)
                {

                    if(temp2->data==removedata)
                    {
                        previous=temp2;
                        delete temp2;
                    }

                    temp2=temp2->link;

                }

            }

        }

    };

    int main()
    {

        linkedlist list;
        list.add(10);
        list.add(100);
        list.add(200);
        list.remove(10);
        list.disp();

    }

显示的输出是：

0
100
200

推荐答案

这与此答案密切相关， a href =http://stackoverflow.com/q/14118775/78845>您的上一个问题。

This is closely related to this answer to your previous question.

让我们格式化此函数并关闭

Let's format this function and take a closer look.

int remove(int removedata)
{
    node* previous;     // Why is this unintialised?
    node* temp2 = head; // Can you think of a more meaningful name for this
                        // variable? Perhaps "current"?
    if (temp2 == NULL)
    {
        exit(0); // Do you really want to exit the program if you try to
                 // remove an item from an empty list? Is there a better way
                 // to handle this?
    }

    if(temp2->link == NULL) // What is so special about a linked list with one element?
    {
        delete temp2; // You've not checked to see if this element is
                      // `removedata`
        head = NULL;  // So calling this function for any list with one element
                      // results in an empty list. This is probably
                      // undesirable
    }
    else
    {
        while (temp2 != NULL) // Consider using a for-loop for iteration. It
                              // couples the increment(s) with the
                              // terminating condition.
        {
            if (temp2->data == removedata) // At this point we have found
                                           // our candidate for removal
            {
                previous = temp2; // This is not the "previous" node, but
                                  // the "current" one. And this variable is
                                  // never used. You do need to know where
                                  // the previous node is and whether it is
                                  // 'head' to update either the 'link' or
                                  // 'head'

                delete temp2;     // This does remove the correct node but
                                  // leaves the previous 'link' pointer
                                  // dangling and any subsequent nodes are
                                  // orphaned.
            }
            temp2 = temp2->link; // See comment above about for-loops.
        }
    }
} // This function is declared `int` but has no return statement.

另一个实现可能看起来像这样（未测试）：

An alternative implementation may look something like this (untested):

void remove(int removedata)
{
    for (node* current = head, * previous = NULL;
        current;
        previous = current, current = current->link)
    {
        if (current->data == removedata)
        {
            if (previous)
            {
                previous->link = current->link;
            }
            else
            {
                head = current->link;
            }
            delete current;
            break;
        }
    }
}

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