我刚刚创建了 LinkedList 的实现(仅用于自学目的。)。我让它运行,但输出结果有点奇怪......这是代码:
#include "stdafx.h"
#include <iostream>
#include <stdio.h>
using namespace std;
template <class T>
class Node{
T datum;
Node<T> *_next;
public:
Node(T datum)
{
this->datum = datum;
_next = NULL;
}
void setNext(Node* next)
{
_next = next;
}
Node* getNext()
{
return _next;
}
T getDatum()
{
return datum;
}
};
template <class T>
class LinkedList{
Node<T> *node;
Node<T> *currPtr;
Node<T> *next_pointer;
int size;
public:
LinkedList(T datum)
{
node = new Node<T>(datum);
currPtr = node; //assignment between two pointers.
next_pointer = node;
size = 1;
}
LinkedList* add(T datum) // return pointer type.
{
Node<T> *temp = new Node<T>(datum);
currPtr->setNext(temp);
currPtr = temp;
size++;
cout<<datum<<" is added.";
return this; //pointer type specification
}
T next()
{
T data = (*next_pointer).getDatum();
cout<<data<<" is visited.";
next_pointer = next_pointer->getNext();
return data;
}
int getSize()
{
return size;
}
};
现在我尝试使用 LinkedList:
int main()
{
LinkedList<int> *list = new LinkedList<int>(1);
list->add(2)->add(3)->add(4);
cout<<endl;
printf("%d %d %d %d",list->next(),list->next(),list->next(),list->next()); \\One
cout<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<endl; \\Two
cout<<list->next()<<endl;\\Three
cout<<list->next()<<endl;
cout<<list->next()<<endl;
cout<<list->next()<<endl;
}
输出一会显示数据:4 3 2 1。二会显示 4 3 2 1。三会显示 1 2 3 4。我不知道运行时发生了什么。所有这些都应该以 1 2 3 4 的顺序输出数据......非常感谢您的帮助!谢谢!