Question

我写的程序是一个拍卖应用程序，到目前为止我所做的是两个构造函数创建了两个ArrayList，它们是绘画和 bid 列表。 Painting列表包含绘画 ID 及其名称。相反，Bid列表包含绘画名称和每个名称的出价金额。

我真正需要的是提示一条消息，在该消息之后用户必须引入其用户名，此时显示绘画名称和最高出价。例如，草莓绘画的最高金额为 1000000 。到目前为止，我已设法显示绘画名称，但不是最高出价。

如果你能帮助我了解从这一点开始从哪里开始，我将非常感激。这已经两天了，我仍然无法找到问题的解决方案。以下是一些代码块：

绘画

的ArrayList构造函数

private ArrayList<Painting> Painting;        
    {
        Painting = new ArrayList<Painting>();
    }

    public void add(int id, String name,String artist,String minibidprice)
    {
        Painting.add (new Painting(id,name,artist,minibidprice));
    }
    public void listAll ()
    {
        for (Painting p : Painting)
        {
            System.out.println(p);
        }
    }

出价列表的ArrayList构造函数

    private ArrayList<Bids>Bids;
    {
        Bids = new ArrayList<Bids>();
    }


    public void add(int id, String bname,String pname,int value)
    {
        Bids.add (new Bids(id,bname,pname,value));
    }
     public void listAll2 ()
    {
        for (Bids b : Bids)
        {
            System.out.println(b);
        }
    }

以下是这些ArrayList个对象的元素。第一个是绘画，下一个是出价。

Bidlists p = new Bidlists();
p.add(1440,"Water-Lillies","Claude Monet","$10,000.00");
p.add(1441,"Flowers in a Blue Vase ","Paul Cezanne","$20,000.00");
p.add(1442,"The Anglers","Henri Rousseau ","$50,000.00");
p.add(1443,"Wild Potato Dreaming","Emily Kngwarreye","$45,000.00");
p.add(1446,"Argenteuil","Claude Monet ","$100,000.00");
p.add(1447,"Luncheon on the Grass","Paul Cezanne","$50,000.00");
p.add(1448,"Antonia","Amadeo Modigliani ","$30,000.00");
p.add(1450,"The Young Apprentice","Amadeo Modigliani","$20,000.00");
p.add(1451,"Woman with Mandolin","Henri Matisse","$15,000.00");
p.add(1457,"Strawberries","Pierre-Auguste Renoir $","$30,000.00");
p.add(1458,"Alhalker Suite","Emily Kngwarreye ","$40,000.00");

Bid lists bid = new Bid lists();
bid.add(1,"Bill Chivers","Wild Potato Dreaming",100000000);
bid.add(2,"Ric Herbert","Strawberries",11000000);
bid.add(2,"Simon","Strawberries",32500000);
bid.add(4,"Bill Chivers","Luncheon on the Grass",90000000);
bid.add(5,"Simon ","Wild Potato Dreaming",110000000);
bid.add(6,"Bill Chivers ","Strawberries",25000000);

Answer 1

您通常会使用Map，因为它旨在允许某些数据与给定密钥相关联（即通过查找）。在这种情况下，您可以将绘画ID映射到最高出价（或者替换为所有出价列表，您按排序顺序维护，因此可以轻松选择最高出价。）

如果由于某种原因必须使用列表，那么您仍然可以通过迭代出价列表来实现您的要求，依次检查每个元素。当您找到与您的绘画ID相关的出价时，您可以考虑它，否则忽略其他绘画的出价。但请注意，与使用Map相比，这在执行速度方面效率较低，并且还涉及更多手动代码（因此更容易出错）。

在选择最高出价方面，如果您给Bid课程一个＆＃34;自然顺序＆＃34;这将有助于很多。（根据出价规模）。这可以通过实现Comparable来完成 - 一旦完成，集合框架就可以为您处理排序实现。但是，您必须将价格实际存储为数字而不是字符串。无论如何，这是一个好主意，因为字符串完全不透明。

一个简单的例子：

// Make Bids comparable
public class Bid implements Comparable<Bid> {
    // fields as before, though we change the amount:
    private final BigDecimal amount;

    // Constructor as before - except you'll need to change the class of the amount

    // now we add the compareTo method
    public int compareTo(Bid other) {
        // This says that Bids are ordered in the opposite way that their amounts
        // are ordered, i.e. highest first
        return -amount.compareTo(other.amount);
    }
}

// Now in your main class you'll need to store the bids for each painting, rather
// than just a list:
private Map<Integer, List<Bid>> bidsByPainting = new HashMap<>();

// Change the method that adds bids so it looks a bit like this:
void add(Bid bid) {
    List<Bid> existing = bidsByPainting.get(bid.getPaintingId());
    if (existing == null) {
        existing = new ArrayList<>();
        bidsByPainting.put(bid.getPaintingId(), existing);
    }
    existing.add(bid);

    // Sort the bids so that the highest is always first.
    // (Note that this only works because we made Bid implement Comparable)
    Collections.sort(existing);
}

// Now, at any time, you can get the highest bid for a painting by calling:
bidsByPainting.get(idYouCareAbout).get(0);

Answer 2

在您的课程中，介绍一种名为

的方法

public Painting getPainting(String paintId)

遍历每个绘图对象，并检查输入arg paintIf是否与您正在迭代的每个绘图对象中的id匹配。
如果发现它返回相同的else，则返回null或抛出异常。

如果你想开发一种有效的方法，那么我会说gor for map with key as id和value as painting object你可以说

map.get(paintId)

获取Paiting对象。

如何比较两个不同数组列表中的对象？

2 个答案: