从一副牌发牌的 Java 代码

Question

我正在尝试编写一个方法来从牌堆顶部删除指定数量的牌并将它们作为数组返回。

这就是我到目前为止所做的。创建牌组、复制牌组的方法、返回数组中指定位置的纸牌的方法、返回纸牌数组大小的方法、洗牌和剪切牌组的方法(不确定这些是否正确)。这是卡片类

public class Card {

    private final int suit; // 0, 1, 2, 3 represent Spades, Hearts, Clubs,
                            // Diamonds, respectively

    private final int value; // 1 through 13 (1 is Ace, 11 is jack, 12 is
                                // queen, 13 is king)

    /*
     * Strings for use in toString method and also for identifying card images
     */
    private final static String[] suitNames = { "s", "h", "c", "d" };
    private final static String[] valueNames = { "Unused", "A", "2", "3", "4",
            "5", "6", "7", "8", "9", "10", "J", "Q", "K" };

    /**
     * Standard constructor.
     * 
     * @param value
     *            1 through 13; 1 represents Ace, 2 through 10 for numerical
     *            cards, 11 is Jack, 12 is Queen, 13 is King
     * @param suit
     *            0 through 3; represents Spades, Hearts, Clubs, or Diamonds
     */
    public Card(int value, int suit) {
        if (value < 1 || value > 13) {
            throw new RuntimeException("Illegal card value attempted.  The "
                    + "acceptable range is 1 to 13.  You tried " + value);
        }
        if (suit < 0 || suit > 3) {
            throw new RuntimeException("Illegal suit attempted.  The  "
                    + "acceptable range is 0 to 3.  You tried " + suit);
        }
        this.suit = suit;
        this.value = value;
    }

    /**
     * "Getter" for value of Card.
     * 
     * @return value of card (1-13; 1 for Ace, 2-10 for numerical cards, 11 for
     *         Jack, 12 for Queen, 13 for King)
     */
    public int getValue() {
        return value;
    }

    /**
     * "Getter" for suit of Card.
     * 
     * @return suit of card (0-3; 0 for Spades, 1 for Hearts, 2 for Clubs, 3 for
     *         Diamonds)
     */
    public int getSuit() {
        return suit;
    }

    /**
     * Returns the name of the card as a String. For example, the 2 of hearts
     * would be "2 of h", and the Jack of Spades would be "J of s".
     * 
     * @return string that looks like: value "of" suit
     */
    public String toString() {
        return valueNames[value] + " of " + suitNames[suit];
    }

    /**
     * [STUDENTS SHOULD NOT BE CALLING THIS METHOD!] Used for finding the image
     * corresponding to this Card.
     * 
     * @return path of image file corresponding to this Card.
     */
    public String getImageFileName() {

        String retValue;
        retValue = suitNames[suit];
        if (value <= 10)
            retValue += value;
        else if (value == 11)
            retValue += "j";
        else if (value == 12)
            retValue += "q";
        else if (value == 13)
            retValue += "k";
        else
            retValue += "Unknown!";
        return "images/" + retValue + ".gif";
    }
}

Deck 方法是我需要帮助的方法

public class Deck {

    private Card[] cards;

    public Deck() {
        cards = new Card[52];
        int numberOfCard = 0;
        for (int suit = 0; suit <= 3; suit++) {
            for (int value = 1; value <= 13; value++) {
                cards[numberOfCard] = new Card(value, suit);
                numberOfCard++;
            }
        }
    }

    public Deck(Deck other) {

        cards = new Card[other.cards.length];
        for (int i = 0; i < other.cards.length; i++) {
            cards[i] = other.cards[i];
        }
    }

    public Card getCardAt(int position) {
        if (position >= cards.length) {
            throw new IndexOutOfBoundsException("Values are out of bounds");
        } else {
            return cards[position];
        }
    }

    public int getNumCards() {
        return cards.length;
    }

    public void shuffle() {
        int temp = 0;
        for (int row = 0; row < cards.length; row++) {
            int random = (int) (Math.random() * ((cards.length - row) + 1));
            Deck.this.cards[temp] = this.getCardAt(row);
            cards[row] = cards[random];
            cards[random] = cards[temp];
        }

    }

    public void cut(int position) {
        Deck tempDeck = new Deck();
        int cutNum = tempDeck.getNumCards() / 2;
        for (int i = 0; i < cutNum; i++) {
            tempDeck.cards[i] = this.cards[52 - cutNum + i];
        }
        for (int j = 0; j < 52 - cutNum; j++) {
            tempDeck.cards[j + cutNum] = this.cards[j];
        }
        this.cards = tempDeck.cards;
    }

    public Card[] deal(int numCards) {
        return cards;
    }

}

Answer 1

如果您不能使用List，那么使用另一个带有牌组顶牌索引的变量似乎是个好主意:

public class Deck {

    private Card[] cards;
    private int topCardIndex;

    public Deck() {
        cards = new Card[52];
        int numberOfCard = 0;
        for(int suit = 0; suit <= 3; suit++){
            for(int value = 1; value <= 13; value++){
                cards[numberOfCard] = new Card(value, suit);
                numberOfCard++;
            }
        }
        topCardIndex = 0;
    }

    public Card getCardAt(int position) {
        if (position >= cards.length - topCardIndex || position < topCardIndex) {
            throw new IndexOutOfBoundsException("Values are out of bounds");
        } else {
            return cards[topCardIndex + position];
        }
    }

    public Card[] deal(int numCards) {
        // FIXME: check bounds, or make method "Card pickCardAt(int position)"
        // that removes the card from the deck
        Card[] drawnCards = new Card[numCards];
        for(int index = 0; index < numCards; index ++) {
            drawnCards[index] = cards[topCard];
            topCard++;
        }
        return drawnCards;
    }

<小时/>

我认为您最好将 Deck 的 card 集合设为 List 而不是 Card[]，这样您就可以进行更好的操作，例如 remove()。

如果您想从牌组中取出前 n 张牌，则只需对结果数组进行 remove() n 次即可(我也会制作一个列表)。

public List<Card> deal(int numCards) {
    List<Card> drawnCards = new ArrayList<Card>(numCards);
    for(int index = 0; index < numCards; index++) {
        drawnCards.add(cards.remove(0));
    }
    return drawnCards;
}