Question

我试图计算列表中某个对象的出现次数，但是看起来它不起作用。

这是我的班级证：

public class Card {
    private int value;
    private String type;
    private String color; 
}

在这里，我正在尝试设置104张卡片组，每张卡片包含2次出现，但看起来我的情况不正确：

public static List<Card> InitializeDeck(){
        for(int i=0;i<104;i++){
            boolean isOk = true;
            while (isOk){
                int col = (int) (Math.floor(Math.random() * 2) + 1);
                if (col == 1) {
                    color = "Black";
                } else {
                    color = "Red";
                }

                value = (int) (Math.floor(Math.random() * 14));
                while (value == 0) {
                    value = (int) (Math.floor(Math.random() * 14));
                }

                int ty = (int) (Math.floor(Math.random() * 4));
                switch (ty) {
                    case 0:
                        type = "Spade";
                        break;
                    case 1:
                        type = "Heart";
                        break;
                    case 2:
                        type = "Diamond";
                        break;
                    case 3:
                        type = "Club";
                        break;
                }
                Card card = new Card(value, type, color);
                if(deck.isEmpty() || deck.stream().filter(line -> card.equals(line)).count()<=1){
                    deck.add(card);
                    isOk=false;
                }
            }

        }

        return deck;
    }

我得到104张纸牌，但是有时同一张纸出现4次，甚至没有一次出现任何提示吗？

Answer 1

我将对您的问题进行一些总结，并给出一个简短的示例，说明如何构建不带随机数和使用枚举的套牌。

首先我们定义枚举：

function foo(){
  console.log(this);
}
foo.call(foo); //foo function
foo.call(); //window object

还有enum Color { RED, BLACK; } enum CardType { SPADE, //as per Yassin's comments you'll probably want to define the type's color later on HEART, //you'd then use HEART(Color.RED) etc. - and provide the correct constructor DIAMOND, CLUB; }类：

Card

并构建甲板：

class Card {
  private final int value;
  private final CardType type;
  private final Color color;

  //I'll omit the constructor, getters, equals and hashcode for simplicity, they should be straight forward
}

计算列表中对象的出现次数

1 个答案: