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C ++链表程序读取字符串

[英]C++ Linked List Program reading strings

Hi right now I am trying to rewrite my code so it can take in each item of the linked list as a string reading a first and last name. 您好,我现在正在尝试重写我的代码,以便它可以将链接列表中的每个项目都当作一个字符串来读取名字和姓氏。 At the moment I got the program to work, but when I enter a first and last name the program treats them as separate elements instead of one single entity. 目前,我可以使用该程序,但是当我输入名字和姓氏时,程序会将它们视为单独的元素,而不是单个实体。 So basically I want "Jack Frost" to have a linked list length of 1 instead 2, and then I try to prompt the user to enter a first and last name so they are deleted from the list. 因此,基本上,我希望“ Jack Frost”的链接列表长度为1而不是2,然后尝试提示用户输入名字和姓氏,以便将其从列表中删除。 Hopefully someone can help. 希望有人可以提供帮助。 I am posting the tester program I made and the header file. 我要发布我制作的测试程序和头文件。 Fair warning the header file is a bit long. 一般警告标头文件有点长。 The tester program needs to print the full list line by line with first and last name. 测试人员程序需要逐行打印完整的列表,包括名字和姓氏。

Update: I do prompt the user to enter first and last name in the same line. 更新:我确实提示用户在同一行中输入名字和姓氏。 They are for some reason being counted as two separate elements, I need to them to be considered on element together. 由于某种原因,它们被视为两个单独的元素,我需要将它们一起考虑在元素上。

#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

This is the tester program below. 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;   

cin input is separated also with space, in your sample code, in the last loop when read cin>>name; 在示例代码中,在读取cin>>name;的最后一个循环中, cin输入也用空格分隔cin>>name; name would take value for all spaces and new line separated words. 名称将使用所有空格和换行符分隔的单词的值。

Use this instead: 使用此代替:

std::string name;

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

Instead of name and surname in two lines get them in one line as following: 代替两行中的名称和姓氏,而是按如下所示将它们放在一行中:

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

To split name and surname copy/paste this code to your solution. 要拆分姓名,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;
}

Then divide name and surname you will write 然后将名称和姓氏分开

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

The easiest solution is to create a structure containing your two strings: 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;
};

Then you can "pass" the structure as the type in your linked list: 然后,您可以将结构作为链接列表中的类型“传递”:

LinkedListType<Person_Name> people;

You may have to adjust the "genericity" or assumptions made by the list or add functionality to the Person_Name to make it conform to the list's interface. 您可能必须调整列表的“通用性”或假设,或向Person_Name添加功能以使其符合列表的界面。

Edit 1: Comparing the data 编辑1:比较数据
The linked list is probably using operator< to order the nodes in the list. 链接列表可能使用operator<对列表中的节点进行排序。 So you will have to add that overloaded operator to your structure: 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;
  }
};

You can implement operator== in a similar fashion. 您可以类似的方式实现operator==

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