java - 尝试修改游戏，以便程序保留未接受的答案列表并在最后打印它们。 (java)

Question

我正在尝试编写一个野餐游戏程序，该程序保留所有被拒绝的项目的列表(但每个项目只出现一次|如果用户输入一个项目两次并且两次都被拒绝，它应该出现仅在拒绝项目列表中出现一次)。游戏结束时，打印出所有被拒绝的项目以及用户的答案被拒绝的次数。我应该使用数组列表。我应该怎么办？这是我到目前为止所拥有的。

import java.util.*;

public class PlayPicnic
{
    public static void main(String[] args)
    {

        Scanner scan = new Scanner(System.in);
        Picnic picnic = new Picnic();

        while (picnic.numberOfItems() < 5)
        {
            System.out.print("What do you want to bring on the picnic? ");
            String item = scan.nextLine();
            if (picnic.okayToBring(item))
            {
                picnic.add(item);
            }
            else
            {
                System.out.println("Sorry, you can't bring " + item);
            }
        }

        System.out.println("\nHere's what we'll have at the picnic:");
        picnic.show();

    }
}

这是相应的野餐类(class)

import java.util.*;

public class Picnic
{
    // INSTANCE VARIABLES:
    private ArrayList<String> stuffToBring; // items to bring on the picnic

    // CONSTRUCTOR:

    //-----------------------------------------------------
    // Construct a new Picnic.
    //-----------------------------------------------------
    public Picnic()
    {
        stuffToBring = new ArrayList<String>(); // initialize list
    }

    //-----------------------------------------------------
    // Given an item s, see if it's okay to add it to the list.
    // Return true if it is, false otherwise:
    //-----------------------------------------------------
    public boolean okayToBring(String s)
    {
        // "Secret rule" -- s can't be an item already in the list:
        if (stuffToBring.contains(s)) // "contains" is in the ArrayList class
        {
            return false;
        }
        else
        {
            return true;
        }
    }

    //-----------------------------------------------------
    // Given an item s, add it to the list (if it's okay to add it)
    //-----------------------------------------------------
    public void add(String s)
    {
        if (okayToBring(s)) 
        {
            stuffToBring.add(s);
        }

    }

    //-----------------------------------------------------
    // Print the items in the list
    //-----------------------------------------------------
    public void show()
    {
        for (int i = 0; i < stuffToBring.size(); i++)
        {
            String s = stuffToBring.get(i);
            System.out.println(s);
        }

        // ALTERNATE APPROACH USING A "for next" LOOP:
        //        for (String s: stuffToBring)
        //        {
        //            System.out.println(s);
        //        }
    }

    //-----------------------------------------------------
    // Returns the number of items in the list:
    //-----------------------------------------------------
    public int numberOfItems()
    {
        return stuffToBring.size();
    }
}

Answer 1

我不确定我是否理解你的问题，但这似乎很简单:添加 ArrayList<String>并插入 item在 else里面while 。然后，在picnic.show()之后，您只需打印 ArrayList .

代码如下:

Scanner scan = new Scanner(System.in);
Picnic picnic = new Picnic();
ArrayList<String> unaccepted = new ArrayList<>();

while (picnic.numberOfItems() < 5)
{
    System.out.print("What do you want to bring on the picnic? ");
    String item = scan.nextLine();
    if (picnic.okayToBring(item))
    {
        picnic.add(item);
    }
    else
    {
        if(!unaccepted.contains(item)) unaccepted.add(item);
        System.out.println("Sorry, you can't bring " + item);
    }
}

System.out.println("\nHere's what we'll have at the picnic:");
picnic.show();
System.out.println(Arrays.toString(unaccepted.toArray()));