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C++ 创建一个每个节点超过 2 个字段的链表

转载 作者:行者123 更新时间:2023-11-27 22:39:36 25 4
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所以我正在尝试创建一个链接列表,它可以为一个富有想象力的图书馆存储书籍。在链表中,每个节点应包含图书馆的分支、作者姓名、书名和该书的拷贝数。

我在创建每个节点有多个字段的链表时遇到困难。我该怎么做才能让每个节点都可以存储 3 个单独的字符串和一个整数,最后存储一个指向下一个节点的指针?

这是我当前的代码。

#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

最佳答案

给定一个结构,例如

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

A LinkedList<Book>将在每个 Node 中包含您想要的所有数据元素.

关于C++ 创建一个每个节点超过 2 个字段的链表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50122521/

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