Question

所以我试图创建一个可以为想象图书馆存储书籍的链接列表。在链表中，每个节点应包含库的分支，作者的姓名，书的标题以及所述书的副本数。

我在创建每个节点有多个字段的链表时遇到了困难。我该如何处理它，以便每个节点可以存储3个单独的字符串和一个整数，最后是一个指向下一个节点的指针？

这是我目前的代码。

#ifndef LINKEDLIST_H
#define LINKEDLIST_H
#include <stdexcept>
using namespace std;

template<typename T>
class Node
{
public:
    T element;  
    Node<T>* next; 

    Node() 
    {
        next = nullptr;
    }

    Node(T element) // Constructor
    {
        this->element = element;
        next = nullptr;
    }
};

template<typename T>
class Iterator : public std::iterator<std::forward_iterator_tag, T>
{
public:
    Iterator(Node<T>* p)
    {
        current = p;
    }

    Iterator operator++() // Prefix ++
    {
        current = current->next;
        return *this;
    }

    Iterator operator++(int dummy) // Postfix ++
    {
        Iterator temp(current);
        current = current->next;
        return temp;
    }

    T& operator*()
    {
        return current->element;
    }

    bool operator==(const Iterator<T>& iterator)
    {
        return current == iterator.current;
    }

    bool operator!=(const Iterator<T>& iterator)
    {
        return current != iterator.current;
    }

private:
    Node<T>* current;
};

template<typename T>

class LinkedList
{
public:
    LinkedList();
    LinkedList(const LinkedList<T>& list);
    virtual ~LinkedList();
    void addFirst(T element);
    void addLast(T element);
    T getFirst() const;
    T getLast() const;
    T removeFirst() throw (runtime_error);
    T removeLast();
    void add(T element);
    void add(int index, T element);
    void clear();
    bool contains(T element) const;
    T get(int index) const;
    int indexOf(T element) const;
    bool isEmpty() const;
    int lastIndexOf(T element) const;
    void remove(T element);
    int getSize() const;
    T removeAt(int index);
    T set(int index, T element);

    Iterator<T> begin() const
    {
        return Iterator<T>(head);
    }

    Iterator<T> end() const
    {
        return Iterator<T>(tail->next);
    }

private:
    Node<T>* head;
    Node<T>* tail;
    int size;
};

template<typename T>
LinkedList<T>::LinkedList()
{
    head = tail = nullptr;
    size = 0;
}

template<typename T>
LinkedList<T>::LinkedList(const LinkedList<T>& list)
{
    head = tail = nullptr;
    size = 0;

    Node<T>* current = list.head;
    while (current != nullptr)
    {
        this->add(current->element);
        current = current->next;
    }
}

template<typename T>
LinkedList<T>::~LinkedList()
{
    clear();
}

template<typename T>
void LinkedList<T>::addFirst(T element)
{
    Node<T>* newNode = new Node<T>(element);
    newNode->next = head;
    head = newNode;
    size++;

    if (tail == nullptr)
        tail = head;
}

template<typename T>
void LinkedList<T>::addLast(T element)
{
    if (tail == nullptr)
    {
        head = tail = new Node<T>(element);
    }
    else
    {
        tail->next = new Node<T>(element);
        tail = tail->next;
    }

    size++;
}

template<typename T>
T LinkedList<T>::getFirst() const
{
    if (size == 0)
        throw runtime_error("Index out of range");
    else
        return head->element;
}

template<typename T>
T LinkedList<T>::getLast() const
{
    if (size == 0)
        throw runtime_error("Index out of range");
    else
        return tail->element;
}

template<typename T>
T LinkedList<T>::removeFirst() throw (runtime_error)
{
    if (size == 0)
        throw runtime_error("No elements in the list");
    else
    {
        Node<T>* temp = head;
        head = head->next;
        if (head == nullptr) tail = nullptr;
        size--;
        T element = temp->element;
        delete temp;
        return element;
    }
}

template<typename T>
T LinkedList<T>::removeLast()
{
    if (size == 0)
        throw runtime_error("No elements in the list");
    else if (size == 1)
    {
        Node<T>* temp = head;
        head = tail = nullptr;
        size = 0;
        T element = temp->element;
        delete temp;
        return element;
    }
    else
    {
        Node<T>* current = head;

        for (int i = 0; i < size - 2; i++)
            current = current->next;

        Node<T>* temp = tail;
        tail = current;
        tail->next = nullptr;
        size--;
        T element = temp->element;
        delete temp;
        return element;
    }
}

template<typename T>
void LinkedList<T>::add(T element)
{
    addLast(element);
}

template<typename T>
void LinkedList<T>::add(int index, T element)
{
    if (index == 0)
        addFirst(element);
    else if (index >= size)
        addLast(element);
    else
    {
        Node<T>* current = head;
        for (int i = 1; i < index; i++)
            current = current->next;
        Node<T>* temp = current->next;
        current->next = new Node<T>(element);
        (current->next)->next = temp;
        size++;
    }
}

template<typename T>
void LinkedList<T>::clear()
{
    while (head != nullptr)
    {
        Node<T>* temp = head;
        head = head->next;
        delete temp;
    }

    tail = nullptr;
    size = 0;
}

template<typename T>
T LinkedList<T>::get(int index) const
{
    if (index < 0 || index > size - 1)
        throw runtime_error("Index out of range");

    Node<T>* current = head;
    for (int i = 0; i < index; i++)
        current = current->next;

    return current->element;
}

template<typename T>
int LinkedList<T>::indexOf(T element) const
{
    // Implement it in this exercise
    Node<T>* current = head;
    for (int i = 0; i < size; i++)
    {
        if (current->element == element)
            return i;
        current = current->next;
    }

    return -1;
}

template<typename T>
bool LinkedList<T>::isEmpty() const
{
    return head == nullptr;
}

template<typename T>
int LinkedList<T>::getSize() const
{
    return size;
}

template<typename T>
T LinkedList<T>::removeAt(int index)
{
    if (index < 0 || index >= size)
        throw runtime_error("Index out of range");
    else if (index == 0)
        return removeFirst();
    else if (index == size - 1)
        return removeLast();
    else
    {
        Node<T>* previous = head;

        for (int i = 1; i < index; i++)
        {
            previous = previous->next;
        }

        Node<T>* current = previous->next;
        previous->next = current->next;
        size--;
        T element = current->element;
        delete current;
        return element;
    }
}

// The functions remove(T element), lastIndexOf(T element),
// contains(T element), and set(int index, T element) are
// left as an exercise

#endif

Answer 1

给定一个结构，例如

struct Book
{
    std::string branch;
    std::string author;
    std::string title;
    int copies;
};

LinkedList<Book>会在每个Node中包含您想要的所有数据元素。

C ++创建每个节点超过2个字段的链接列表

1 个答案: