java - 如何让程序连续处理方法

问题描述

我的问题是：为什么我的程序在处理文本文件中的行时不重复。

我正在编写一个管理车库的类-车库中的汽车被放入不超过 7 个的堆栈中，如果已满，它们将进入不超过 5 个的队列。有手动添加汽车的方法和一种用于读取文本文件的输入。我还获得了一个 stackInt 类，我被要求在创建堆栈对象时使用它。它只有 4 种方法，这让事情变得很麻烦：

public interface StackInt<E> {
    E push(E obj);
    E peek();
    E pop();
    boolean empty();
}

所以我不得不用一些方法来编写我的代码来规避没有其他有用的堆栈方法。我的大多数方法都按预期工作：它读取文本文件并对其进行处理，但是如果例如从车库中移除一辆车，则应该插入队列中的另一辆车，它不会重复该过程循环。这是我的代码：

import java.util.*;
import java.util.LinkedList;
import java.io.FileNotFoundException; 
import java.util.Scanner;
import java.io.File;

public class Garage{
   
   public String line;
   public String stackToString;
   public int position;
   public int numAdded;
   LinkedList<String> queue = new LinkedList<>();
   StackInt<String> stack = new LinkedStack<>();

   public Garage(){
   }
   
   public Garage(String fileName){
      try{
         File file = new File(fileName);
         Scanner scan = new Scanner(file);
         while(scan.hasNextLine()) {
          line = scan.nextLine();
          String[] data = line.split(" ");
          if (data[0].equals("a"))
              arrival(data[1]);
          else if (data[0].equals("d"))
              departure(data[1]);
         }
       }  
      catch(FileNotFoundException e) {
         System.out.println("File not found");
      }
   }
       
   public boolean arrival(String license){
     boolean added = false;
     
     if(numAdded < 7){
         stack.push(license);
         added = true;
     } else if (queue.size() < 5) {
          added = queue.add(license);
          numAdded--;
       }
     if(added){
       numAdded++;
      }
  
      return true;
  }       
   
   public int departure(String license){
     Stack<String> temp = new Stack();
     
     while(!stack.empty()){
         temp.push(stack.pop());
         
     }
     position = temp.indexOf(license);
         temp.remove(license);
     while(!temp.isEmpty()){
         stack.push(temp.pop());
     }
     return position;    
   }
      
   public int numberParked(){
      return numAdded;
   }
   
   public int numberWaiting(){
      return queue.size();
   }
   
   public String toString(){                         
      Stack<String> tempStack = new Stack();
      while (!stack.empty()){
        tempStack.push(stack.pop()); 
        stackToString = tempStack.toString().replace("[", "").replace("]", ""); 
      } 
      while (!tempStack.empty()){
        stack.push(tempStack.pop()); 
      }     
      return "Cars in Garage: " + stackToString + "\n" + "Cars waiting: " + (queue.toString().replace("[", "").replace("]", ""));
  } 
}

我正在测试的是：

public class GarageTest
{
    public static void main (String [] args) 
   {
      Garage g1 = new Garage("parking.txt");
      System.out.println("Number parked: " + g1.numberParked());
      System.out.println("Number waiting: " + g1.numberWaiting());
      System.out.println("Parking WEB445 ... " + g1.arrival("WEB445"));
      System.out.println("Parking BEA345 ... " + g1.arrival("BEA345"));
      System.out.println(g1);
      System.out.println("Z23YTU departs after " + g1.departure("B12GFT") + " car(s) moved");
      System.out.println(g1);
    }
}

输出是：

 Number parked: 7
    Number waiting: 5
    Parking WEB445 ... true
    Parking BEA345 ... true
    Cars in Garage: Y23456, X12345, B12GFT, Z23YTU
    Cars waiting: W321RE, CVBNMK, DFGHJK, ERTYUI, FGHJKL
    Z23YTU departs after 2 car(s) moved
    Cars in Garage: Y23456, X12345, Z23YTU
    Cars waiting: W321RE, CVBNMK, DFGHJK, ERTYUI, FGHJKL

“车库里的汽车”线的预期是：

DFGHJK CVBNMK R23EWQ W321RE Y23456 X12345 B12GFT Z23YTU 
with the queue on the next line. It's removing a few that have "d", but not filling the garage 
afterward

正在处理的文本文件（a 表示到达，d 表示离开）：

a A123TR
a Z23YTU
a Z23YTU
a ERW345
d ERW345
a B12GFT
d a23TR
a X12345
a Y23456
a W321RE
d R23EWQ
a CVBNMK
a DFGHJK
a ERTYUI
a FGHJKL
a GHJKL9
a HJKL98

标签： java

java - 如何让程序连续处理方法

问题描述

解决方案

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