java - 将 int 值赋给字符串

Question

我正在设计一个 Black Jack 程序。我的问题是，我使用开关盒生成随机卡。但是当需要比较卡片的值(value)时......让我们说 if(pCard1 + pCard2 > 21) 它不允许我比较它们，因为它们是字符串。但是，它们必须是字符串，因为我必须能够分配 String "Queen" 但我还需要 "Queen"的 int 值为 10。我不知道该怎么做这个。

这是我目前所拥有的。我没有包括我的整个代码，因为它有几百行长。仅包括二十一点部分。

System.out.println("---------------------Black Jack--------------------------");
    System.out.println("Welcome to BlackJack!");
    System.out.println("Available balance is $"+balance);
    System.out.print("How much would you like to bet on this hand?: ");
    int bet = input.nextInt();
    balance -= bet;
    System.out.println("You just bet $"+bet+"......Dealing cards!");
    System.out.println("----------------------------------------------------------");

    String pCard1 = dealCard();
    String pCard2 = dealCard();

    System.out.println("Your hand is a "+pCard1+" and "+pCard2);
    System.out.print("Would you like to Hit or Stand? :");
    String hOs = input.next();

    if(hOs.equals("Hit")) {
        String pCard3 = dealCard();
        if(pCard1 + pCard2 + pCard3);
    }
    }
}
public static String dealCard() {
    int value = (int)(Math.random() * 13);
    String returnString = "";
    switch ( value ) {
        case 1:   returnString = "Ace"; break;
        case 2:   returnString = "2"; break;
        case 3:   returnString = "3"; break;
        case 4:   returnString = "4"; break;
        case 5:   returnString = "5"; break;
        case 6:   returnString = "6"; break;
        case 7:   returnString = "7"; break;
        case 8:   returnString = "8"; break;
        case 9:   returnString = "9"; break;
        case 10:  returnString = "10"; break;
        case 11:  returnString = "Jack"; break;
        case 12:  returnString = "Queen"; break;
        case 13:  returnString = "King"; break;
    }
    return returnString;
  }
}    

Answer 1

我认为枚举是解决此类问题的最佳方式。

如果我们像这样为 Suit 和 Pips 定义枚举

enum Suit {
    Hearts, Diamonds, Clubs, Spades;
}

enum Pips {
    Ace, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King;
}

如果我们定义一个类 Card:

class Card {
    Suit suit;
    Pips pips;

    public Card(Suit suit, Pips pips) {
        this.suit = suit;
        this.pips = pips;
    }

    @Override
    public String toString() {
        return pips.toString() + " of " + suit.toString();
    }
}

Card 使用 toString() 将 Suit 和 Pips 转换为字符串，而零索引数值可以使用 ordinal():

    Card card = new Card(Suit.Spades, Pips.Ace);
    System.out.println("Card: " + card);
    System.out.println("Numeric suit: " + card.suit.ordinal());
    System.out.println("Numeric pips: " + card.pips.ordinal());

这个简单的实现有几个限制。首先，花色或点数的名称只能是枚举名称的字符串版本。如果您希望所有枚举都采用大写形式，那会怎样？第二个问题是序数总是零索引的。如果您需要一些其他的 id 系统怎么办？

在这种情况下，我们将映射添加到每个枚举以保存这些映射。不幸的是，我们不能使用继承在枚举之间共享代码，因此样板变得有点冗长。

enum Suit {
    HEARTS(1, "Hearts"), //
    DIAMONDS(2, "Diamonds"), //
    CLUBS(3, "Clubs"), //
    SPADES(4, "Spades");

    private static Map<String, Suit> nameToEnum = new HashMap<>();
    private static Map<Integer, Suit> idToEnum = new HashMap<>();

    static {
        for (Suit suit : Suit.values()) {
            nameToEnum.put(suit.getName(), suit);
            idToEnum.put(suit.getId(), suit);
        }
    }

    private int id;
    private String name;

    private Suit(int id, String name) {
        this.id = id;
        this.name = name;
    }

    public int getId() {
        return id;
    }

    public String getName() {
        return name;
    }

    public static Suit fromName(String name) {
        return nameToEnum.get(name);
    }

    public static Suit fromId(int id) {
        return idToEnum.get(id);
    }
}

enum Pips {
    ACE(1, 1, "Ace"), //
    TWO(2, 2, "Two"), //
    THREE(3, 3, "Three"), //
    FOUR(4, 4, "Four"), //
    FIVE(5, 5, "Five"), //
    SIX(6, 6, "Six"), //
    SEVEN(7, 7, "Seven"), //
    EIGHT(8, 8, "Eight"), //
    NINE(9, 9, "Nine"), //
    TEN(10, 10, "Ten"), //
    JACK(11, 10, "Jack"), //
    QUEEN(12, 10, "Queen"), //
    KING(13, 10, "King");

    private static Map<String, Pips> nameToEnum = new HashMap<>();
    private static Map<Integer, Pips> idToEnum = new HashMap<>();

    static {
        for (Pips pips : Pips.values()) {
            nameToEnum.put(pips.getName(), pips);
            idToEnum.put(pips.getId(), pips);
        }
    }

    private int id;
    private int value;
    private String name;

    private Pips(int id, int value, String name) {
        this.id = id;
        this.value = value;
        this.name = name;
    }

    public int getId() {
        return id;
    }

    public int getValue() {
        return value;
    }

    public String getName() {
        return name;
    }

    public static Pips fromName(String name) {
        return nameToEnum.get(name);
    }

    public static Pips fromId(int id) {
        return idToEnum.get(id);
    }

}

使用这些新枚举，我们修改 Card.toString() 以使用新的 getName() 方法:

    return pips.getName() + " of " + suit.getName();

现在，当我们使用卡片时，我们可以使用任何 ID、名称或枚举，并在它们之间自由转换:

    Card card = new Card(Suit.SPADES, Pips.ACE);
    System.out.println("Card: " + card);
    System.out.println("Numeric suit: " + card.suit.getId());
    System.out.println("Numeric pips: " + card.pips.getId());
    System.out.println("Suit by ID: " + Suit.fromId(3).getName());
    System.out.println("Pips by Name: " + Pips.fromName("Ace").getName());
    System.out.println("Pips value by Name: " + Pips.fromName("King").getValue());