c++ - 我的链表方法有什么改进吗？

Question

我一直在编写一个程序来提高我对链表及其功能的了解。我想知道你们中的一些人是否可以查看我的代码并注意到您可能熟悉的任何错误或任何一般错误。这些函数适用于我的测试函数，但显然，我没有测试所有可能的场景。

    // LinkedList.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>

using std::cout;
using std::cin;
using std::endl;

struct node
{
    int value;
    node* next;
};

node* push(node*, int);
node* pop(node*);
node* pop(node*, int);
node* insert(node*, int, int);
void printList(const node*);

int _tmain(int argc, _TCHAR* argv[])
{
    node* head = NULL;

    for(int i = 1; i <= 15; ++i)
    {
        head = push(head, i);
    }

    for(int j = 14; j >= 0; j = j - 2)
    {
        head = pop(head, j);
        printList(head);
    }

    head = pop(head);
    head = insert(head, 4, 27);
    printList(head);
    cin.ignore();
}

node* push(node* read, int val)
{
    node* write = read;
    if(read == NULL)
    {
        read = new node;
        read->next = NULL;
        read->value = val;
        cout << "Node Head: " << read->value << endl;
        return read;
    }
    else
    {
        while(read->next != NULL)
        {
            read = read->next;
        }
        read->next = new node;
        read->next->next = NULL;
        read->next->value = val;
        read = read->next;
        cout << "Node Link: " << read->value << endl;
        return write;
    }
}

node* pop(node* read)
{
    node* write = read;
    if(read->next == NULL)
    {
        delete read;
        read = NULL;
        return write;
    }
    else
    {
        while(read->next != NULL)
        {
            if(read->next->next == NULL)
            {
                cout << "Pop: " << read->next->value << endl;
                delete read->next;
                read->next = NULL;
            }
            else
            {
                read = read->next;
            }
        }
        return write;
    }
}

node* pop(node* read, int pos)
{
    node* write = read;
    if(read->next == NULL)
    {
        delete read;
        return write;
    }
    else
    {   
        if(pos == 0)
        {
            node* old = read;
            cout << "Pop: " << read->value << endl;
            read = read->next;
            delete old;
            return read;
        }
        else
        {
            for(int i = 0; i < (pos - 1); i++)
            {
                read = read->next;
            }
            node* old = read->next;
            cout << "Pop: " << old->value << endl;
            read->next = read->next->next;
            delete old;
            return write;
        }
    }
}

node* insert(node* read, int pos, int val)
{
    node* write = read;
    for(int i = 0; i < (pos - 1); i++)
    {
        read = read->next;
    }
    node* ins = new node;
    ins->value = val;
    cout << "Insert: " << ins->value << endl;
    ins->next = read->next;
    read->next = ins;
    return write;
}

void printList(const node* read)
{
    if(read != NULL)
    {
        cout << "List Item: " << read->value << endl;
        printList(read->next);
    }
}

/****************OUTPUT*********************
Node Head: 1
Node Link: 2
Node Link: 3
Node Link: 4
Node Link: 5
Node Link: 6
Node Link: 7
Node Link: 8
Node Link: 9
Node Link: 10
Node Link: 11
Node Link: 12
Node Link: 13
Node Link: 14
Node Link: 15
Pop: 15
List Item: 1
List Item: 2
List Item: 3
List Item: 4
List Item: 5
List Item: 6
List Item: 7
List Item: 8
List Item: 9
List Item: 10
List Item: 11
List Item: 12
List Item: 13
List Item: 14
Pop: 13
List Item: 1
List Item: 2
List Item: 3
List Item: 4
List Item: 5
List Item: 6
List Item: 7
List Item: 8
List Item: 9
List Item: 10
List Item: 11
List Item: 12
List Item: 14
Pop: 11
List Item: 1
List Item: 2
List Item: 3
List Item: 4
List Item: 5
List Item: 6
List Item: 7
List Item: 8
List Item: 9
List Item: 10
List Item: 12
List Item: 14
Pop: 9
List Item: 1
List Item: 2
List Item: 3
List Item: 4
List Item: 5
List Item: 6
List Item: 7
List Item: 8
List Item: 10
List Item: 12
List Item: 14
Pop: 7
List Item: 1
List Item: 2
List Item: 3
List Item: 4
List Item: 5
List Item: 6
List Item: 8
List Item: 10
List Item: 12
List Item: 14
Pop: 5
List Item: 1
List Item: 2
List Item: 3
List Item: 4
List Item: 6
List Item: 8
List Item: 10
List Item: 12
List Item: 14
Pop: 3
List Item: 1
List Item: 2
List Item: 4
List Item: 6
List Item: 8
List Item: 10
List Item: 12
List Item: 14
Pop: 1
List Item: 2
List Item: 4
List Item: 6
List Item: 8
List Item: 10
List Item: 12
List Item: 14
Pop: 14
Insert: 27
List Item: 2
List Item: 4
List Item: 6
List Item: 8
List Item: 27
List Item: 10
List Item: 12
*******************************************/

Answer 1

嗯，首先，对于一般用途，您应该使用标准库:std:vector ，或者，如果你真的需要一个链表，std::list<>.但是，由于这是自学练习，我们将跳过它。

这给我们带来了下一个问题:作为教学练习，它确实不是生产代码，那么我们应该提示什么？ cout 在函数中间？？

我看到的一个特殊问题是这样的代码:

    read = new node;    
    read->next = NULL;    
    read->value = val; 

节点对象的构造函数应该处理将其设置为 next成员为空。就此而言，它还应该处理设置值，因此这段代码实际上应该是:

    read = new node(val);

另一个问题:在 pop() 中，您有:

node* write = read; 
if(read->next == NULL) 
{ 
    delete read; 
    read = NULL; 
    return write; 
} 

设置read为 null 是没有意义的——它是一个即将超出范围的局部变量。而你正在返回 write , 等于 read , 已被删除。

此外，您在代码中几乎没有使用 C++ 和面向对象编程的任何功能:如果我们忽略 cout，它基本上只是通过 new 和 delete 分配内存的 C 代码。正如我所指出的，它可以从构造函数中受益匪浅。析构函数可能也很有用，再加上一个 List 类，它将包含头节点，并将您的所有函数作为成员。