排序后将数据添加到链接列表

时间:2017-10-14 20:09:18

标签: c++ sorting linked-list

我正在尝试读取一个文本文件,根据文本文件中的数字对每个用户进行排序,然后将其添加到链接列表中。然后再次按降序显示每个节点的用户编号。我也是尝试添加编辑功能,例如删除或更新内容。

txt文件就像

John Doe; 10

莎莉莉莉; 5

詹姆斯沃森; 12

我所取得的成就是:

list.h:

#include<string>
#ifndef LIST_H
#define LIST_H
class list{
private:
    struct node{
        std::string  data;
        node* next;
    };
    typedef struct node* nodePtr;
    nodePtr head;
    nodePtr curr;
    nodePtr temp;
 public:
    list();
    void AddNode(std::string addData);
    void DeleteNode(std::string delData);
    void PrintList();
 };
#endif

list.cpp

#include <cstdlib>
#include <iostream>
#include "list.h"
using namespace std;
list::list(){ //constructor
    head = NULL;
    curr= NULL;
    temp=NULL;  
}
   void list::AddNode(string addData){
    nodePtr n = new node; //nodePtr is node*
    n->next=NULL; //find node n is pointing to, access its next element make it point to null
    n->data= addData;
    if(head != NULL) { // if we have at least 1 element in the list .
    curr = head; // take the current pointer we are working with and make it same with head pointer pointing to.(current= front of list)
    while(curr->next !=NULL){ // are we at the end of the list.
    curr = curr->next;//we are not end of the list.curr pointer points next node.
    }
    curr->next = n; 
    }else{ //if we dont have at least 1 element in the list.
        head =n;
    }
}

void list::DeleteNode(string delData){
    nodePtr delPtr = NULL;
    temp = head;
    curr = head;
    while(curr != NULL && curr->data != delData){ // look for data user wants to delete
    temp = curr;
    curr =  curr->next;
    }
    if(curr == NULL){ // we cant find we are looking for.
    cout << delData << "not in list"<<endl;
    delete delPtr;
    }else{ // we found it
        delPtr = curr;
        curr = curr->next; // with those 2 lines we are patching the hole in the list.
        temp->next = curr;
        if(delPtr == head){
        head = head->next;
        temp = NULL;
        }
        delete delPtr;
        cout<<delData<<"deleted"<<endl;
    }
}

 void list::PrintList(){
    curr = head;
    while(curr !=NULL){
        cout<<curr->data<<endl;
        curr = curr->next; 
    }
}

int main(){
    list mylist;
    mylist.AddNode("hello");
    mylist.AddNode("how u doin");
    mylist.AddNode("good");
    mylist.PrintList();
    return 0;
}

阅读txt文件后,我应该将它们放入一个向量(或直接列表)然后排序然后放到链表?我想它错了我应该直接把它们排序然后在屏幕上打印之前排序。我也不知道如何通过用户输入功能更新节点。

更新: 我实现了从txt文件向链表添加内容。

string file;
    ifstream filein;


    cout << "Enter file name:"<<endl;

    cin >> file;

    filein.open(file);
    for(;filein.fail();)
    {
        cout << "Cannot open the file"<<endl;
        cout << "Enter file name:"<<endl;

        cin >> file;

        filein.open(file);
    }

        string cline;
        string cname;
        string csurname;
        int money;
        string smoney;
        string lastdata;
        char delimiter=';';

        while (std::getline(filein, cline)){
                std::istringstream iss(cline);

                while(iss >> cname >> csurname >> delimiter >> money){


                    ostringstream temp; // int to string
                    temp<<money;
                    smoney=temp.str();

                    lastdata = cname+" "+csurname+" "+smoney;
                    mylist.AddNode(lastdata);







                }







        }

        mylist.PrintList();

现在它增加了喜欢 约翰多伊10 莎莉莉莉5 问题是我将如何达到这个10并在屏幕上打印时对它们进行排序

1 个答案:

答案 0 :(得分:0)

解决问题的一种方法是创建两种方法,一种处理比较,另一种处理链表。

以下方法涉及比较。在这种情况下,它比较字符串的长度,但它可以是任何东西。

bool list::IsGreater(nodePtr a, nodePtr b)
{
   if (a->data.length() > b->data.length())
      return true;
   return false;
}

第二种方法对链表进行排序。即使存在排序方法,最简单的方法是在方法AddNode期间对列表进行排序。

void list::SortList() {
   for (nodePtr i = head; i != NULL; i = i->next) {
      nodePtr prev = NULL;
      for (nodePtr j = head; j != NULL && j->next != NULL; j = j->next) {
         if (IsGreater(j, j->next)) {
            if (prev)
               prev->next = j->next;
            else
               head = j->next;

            nodePtr tmp = j->next;
            j->next = j->next->next;
            tmp->next = j;
            j = tmp;
         }

         if (prev == NULL)
            prev = head;
         else
            prev = j;
      }

      prev = i;
   }
}