C++链表程序读取字符串

Question

嗨，现在我正在尝试重写我的代码，以便它可以将链表的每个项目作为读取名字和姓氏的字符串接收。目前我让程序工作，但是当我输入名字和姓氏时，程序将它们视为单独的元素而不是一个单一的实体。所以基本上我希望“Jack Frost”的链表长度为 1 而不是 2，然后我尝试提示用户输入名字和姓氏，以便将它们从列表中删除。希望有人可以提供帮助。我正在发布我制作的测试程序和头文件。公平警告头文件有点长。测试程序需要用名字和姓氏逐行打印完整的列表。

更新：我确实提示用户在同一行中输入名字和姓氏。由于某种原因，它们被视为两个独立的元素，我需要将它们放在一起考虑。

#ifndef H_LinkedListType
#define H_LinkedListType


#include <iostream>

                            //Definition of the node

template <typename Type>
struct nodeType
{
    Type info;
    nodeType<Type> *link;
};

template<typename Type>
class linkedListType
{
  public:
    const linkedListType<Type>& operator=(const linkedListType<Type>&); 
                                                               //Overload the assignment operator
    void initializeList(); 
                                   //Initialize the list to an empty state
                                   //Post: first = NULL, last = NULL
    bool isEmptyList();
                                   //Function returns true if the list is empty;
                                   //otherwise, it returns false
    bool isFullList();
                                   //Function returns true if the list is full;
                                   //otherwise, it returns false
    void print();
                                   //Output the data contained in each node
                                   //Pre: List must exist
                                   //Post: None
    int length();
                                   //Return the number of elements in the list
    void destroyList();
                                   //Delete all nodes from the list
                                   //Post: first = NULL, last = NULL
    void retrieveFirst(Type& firstElement); 
                                   //Return the info contained in the first node of the list
                                   //Post: firstElement = first element of the list
    void search(const Type& searchItem);
                                   //Outputs "Item is found in the list" if searchItem is in 
                                   //the list; otherwise, outputs "Item is not in the list"
    void insertFirst(const Type& newItem);
                                   //newItem is inserted in the list
                                   //Post: first points to the new list and the 
                                   //       newItem inserted at the beginning of the list
    void insertLast(const Type& newItem);
                                   //newItem is inserted in the list
                                   //Post: first points to the new list and the 
                                   //     newItem is inserted at the end of the list
                                   //     last points to the last node in the list
    void deleteNode(const Type& deleteItem);
                                   //if found, the node containing deleteItem is deleted 
                                   //from the list
                                   //Post: first points to the first node and
                                   //  last points to the last node of the updated list
    linkedListType(); 
                                   //default constructor
                                   //Initializes the list to an empty state
                                   //Post: first = NULL, last = NULL 
    linkedListType(const linkedListType<Type>& otherList); 
                                   //copy constructor
    ~linkedListType();   
                                   //destructor
                                   //Deletes all nodes from the list
                                   //Post: list object is destroyed 

  protected:
    nodeType<Type> *first;                             //pointer to the first node of the list
    nodeType<Type> *last;                              //pointer to the last node of the list 
};

template<typename Type>
void linkedListType<Type>::initializeList()
{
    destroyList();                             //if the list has any nodes, delete them
}

template<typename Type>
void linkedListType<Type>::print()
{
    nodeType<Type> *current;                             //pointer to traverse the list

    current = first;                               //set current so that it points to 
                                                   //the first node
    while(current != NULL)                             //while more data to print
    {
       cout<<current->info<<" ";
       current = current->link;
    }
}                            //end print

template<typename Type>
int linkedListType<Type>::length()
{
    int count = 0;
    nodeType<Type> *current;                             //pointer to traverse the list

    current = first;

    while (current!= NULL)
      {
       count++;
       current = current->link;
    }

    return count;
}                              // end length

template<typename Type>
void linkedListType<Type>::search(const Type& item)
{
    nodeType<Type> *current;                             //pointer to traverse the list
    bool found;

    if(first == NULL)                              //list is empty
        cout<<"Cannot search an empty list. "<<endl;
    else
    {
        current = first;                              //set current pointing to the first 
                                                      //node in the list

        found = false;                                //set found to false

        while(!found && current != NULL)                             //search the list
            if(current->info == item)                                  //item is found
                found = true;
            else
                current = current->link;                             //make current point to 
                                                                     //the next node

        if(found)
            cout<<"Item is found in the list."<<endl;
        else
            cout<<"Item is not in the list."<<endl;
   }                             //end else
}                            //end search

template<typename Type>
void linkedListType<Type>::insertLast(const Type& newItem)
{
    nodeType<Type> *newNode;                             //pointer to create the new node

    newNode = new nodeType<Type>;                             //create the new node
    newNode->info = newItem;                                  //store the new item in the node
    newNode->link = NULL;                                     //set the link field of new node
                                                              //to NULL

    if(first == NULL)                              //if the list is empty, newNode is 
                                                //both the first and last node
    {
        first = newNode;
        last = newNode;
    }
    else                                 //if the list is not empty, insert newNnode after last
    {
        last->link = newNode;                             //insert newNode after last
        last = newNode;                             //make last point to the actual last node
    }
}                            //end insertLast

template<typename Type>
void linkedListType<Type>::deleteNode(const Type& deleteItem)
{
    nodeType<Type> *current;                             //pointer to traverse the list
    nodeType<Type> *trailCurrent;                             //pointer just before current
    bool found;

    if(first == NULL)                                //Case 1; list is empty. 
        cout<<"Can not delete from an empty list.\n";
    else
    {
        if(first->info == deleteItem)                             //Case 2 
        {
            current = first;
            first = first ->link;
            if(first == NULL)                                //list had only one node
                last = NULL;
            delete current;
        }
        else                              //search the list for the node with the given info
        {
            found = false;
            trailCurrent = first;                               //set trailCurrent to point to
                                                                //the first node
            current = first->link;                              //set current to point to the 
                                                                //second node

            while((!found) && (current != NULL))
            {
                if(current->info != deleteItem) 
                {
                    trailCurrent = current;
                    current = current-> link;
                }
                else
                    found = true;
            }                                           // end while

            if(found)                                                   //Case 3; if found, delete the node
            {
                trailCurrent->link = current->link;

                if(last == current)                                     //node to be deleted was 
                                                                        //the last node
                    last = trailCurrent;                                //update the value of last

                delete current;                                         //delete the node from the list
            }
            else
                cout<<"Item to be deleted is not in the list."<<endl;
        }                             //end else
    }                             //end else
}                             //end deleteNode

#endif

这是下面的测试程序。

//This program tests various operation of a linked list

#include "stdafx.h"
#include <string>//Need to import string to the class to be able to use that type
#include <iostream>
#include "linkedList.h"

using namespace std;

int main()
{
    linkedListType<string> list1, list2;                
    int num;
    string name;//Input from the user


    //cout<<"Enter numbers ending with -999"
        //<<endl;                                       
    //cin>>num; 

    cout<<"Please enter first and last name in each line, enter end to close program" //Prompting the user to enter first and last name on each line til they specify "end"
        <<endl;
        cin>>name;

    while(name != "end")//Checks if user entered name if not it will proceed to keep prompting the user                                 
    {
        list1.insertLast(name);//inserts the name at the end of the list                        
        cin>>name;                                  
    }

    cout<<endl;                                     

    cout<<"List 1: ";                       
    list1.print();//Prints all the names line by line                                   
    cout<<endl;                                     
    cout<<"Length List 1: "<<list1.length()//Gives length of how many names are in the list 
        <<endl;                                     

    list2 = list1;                                  //test the assignment operator 

    cout<<"List 2: ";                       
    list2.print();                                  
    cout<<endl;                                     
    cout<< "Length List 2: "<< list2.length()   <<endl;                                     

    cout<< "Enter the name to be "  << "deleted: ";                             
    cin>>num;                                       
    cout<<endl;                                     

    list2.deleteNode(name);                         

    cout<<"After deleting the node, "
        << "List 2: "<<endl;                            
    list2.print();                                  
    cout<<endl;                                     

    cout<<"Length List 2: "<<list2.length()
        <<endl;                                     

    system("pause");

    return 0;   

Answer 1

cin 输入也用空格分隔，在您的示例代码中，在读取 cin>>name; 时的最后一个循环中，名称将对所有空格和换行符进行分隔。

改用这个：

std::string name;

std::getline(std::cin, name);

Answer 2

最简单的解决方案是创建一个包含两个字符串的结构：

struct Person_Name
{
  std::string first;
  std::string last;
};

然后您可以将结构“传递”为链表中的类型：

LinkedListType<Person_Name> people;

您可能需要调整列表的“通用性”或假设，或者向 Person_Name 添加功能以使其符合列表的界面。

编辑 1：比较数据
链表可能使用operator< 对列表中的节点进行排序。因此，您必须将重载运算符添加到您的结构中：

struct Person_Name
{
  std::string first_name;
  std::string second_name;
  bool operator<(const& Person_Name pn) const
  {
    if (second_name == pn.second_name)
    {
       return first_name < pn.first_name;
    }
    return second_name < pn.second_name;
  }
};

您可以以类似的方式实现operator==。

Answer 3

代替两行中的姓名和姓氏，将它们放在一行中，如下所示：

char name[256]
std::cout << "Please, enter your name: and surname";
std::cin.getline (name,256);

要拆分姓名和姓氏，请将此代码复制/粘贴到您的解决方案中。

std::vector<std::string> &split(const std::string &s, char delim, std::vector<std::string> &elems) {
    std::stringstream ss(s);
    std::string item;
    while (std::getline(ss, item, delim)) {
        elems.push_back(item);
    }
    return elems;
}

std::vector<std::string> split(const std::string &s, char delim) {
    std::vector<std::string> elems;
    split(s, delim, elems);
    return elems;
}

然后把你要写的名字和姓氏分开

vector<string> person = split(name, ' ');
person.at(0)  // the name of the person 
person.at(1)  //the surname of the person