Question

我正在为我的计算机科学课开展一个项目。我的老师希望我们创建一个程序，这个程序将有一副牌，能够将它们洗牌并交易。他希望我们创建一个具有变量suitName，suitValue，faceName和faceValue的卡片类。我目前的代码如下：

//Construct Array
Card[] deck = new Card[52];
//Define cards
Card aceSpades = new Card("Spades", 1, "Ace", 1);
Card aceHearts = new Card("Hearts", 2, "Ace", 1);
Card aceClubs = new Card("Clubs", 3, "Ace", 1);
Card aceDiamonds = new Card("Diamonds", 4, "Ace", 1);
Card twoSpades = new Card("Spades", 1, "Two", 2);
Card twoHearts = new Card("Hearts", 2, "Two", 2);
Card twoClubs = new Card("Clubs", 3, "Two", 2);
Card twoDiamonds = new Card("Diamonds", 4, "Two", 2);
Card threeSpades = new Card("Spades", 1, "Three", 3);
Card threeHearts = new Card("Hearts", 2, "Three", 3);
Card threeClubs = new Card("Clubs", 3, "Three", 3);
Card threeDiamonds = new Card("Diamonds", 4, "Three", 3);
Card fourSpades = new Card("Spades", 1, "Four", 4);
Card fourHearts = new Card("Hearts", 2, "Four", 4);
Card fourClubs = new Card("Clubs", 3, "Four", 4);
Card fourDiamonds = new Card("Diamonds", 4, "Four", 4);
Card fiveSpades = new Card("Spades", 1, "Five", 5);
Card fiveHearts = new Card("Hearts", 2, "Five", 5);
Card fiveClubs = new Card("Clubs", 3, "Five", 5);
Card fiveDiamonds = new Card("Diamonds", 4, "Five", 5);
Card sixSpades = new Card("Spades", 1, "Six", 6);
Card sixHearts = new Card("Hearts", 2, "Six", 6);
Card sixClubs = new Card("Clubs", 3, "Six", 6);
Card sixDiamonds = new Card("Diamonds", 4, "Six", 6);
Card sevenSpades = new Card("Spades", 1, "Seven", 7);
Card sevenHearts = new Card("Hearts", 2, "Seven", 7);
Card sevenClubs = new Card("Clubs", 3, "Seven", 7);
Card sevenDiamonds = new Card("Diamonds", 4, "Seven", 7);
Card eightSpades = new Card("Spades", 1, "Eight", 8);
Card eightHearts = new Card("Hearts", 2, "Eight", 8);
Card eightClubs = new Card("Clubs", 3, "Eight", 8);
Card eightDiamonds = new Card("Diamonds", 4, "Eight", 8);
Card nineSpades = new Card("Spades", 1, "Nine", 9);
Card nineHearts = new Card("Hearts", 2, "Nine", 9);
Card nineClubs = new Card("Clubs", 3, "Nine", 9);
Card nineDiamonds = new Card("Diamonds", 4, "Nine", 9);
Card tenSpades = new Card("Spades", 1, "Ten", 10);
Card tenHearts = new Card("Hearts", 2, "Ten", 10);
Card tenClubs = new Card("Clubs", 3, "Ten", 10);
Card tenDiamonds = new Card("Diamonds", 4, "Ten", 10);
Card jackSpades = new Card("Spades", 1, "Jack", 11);
Card jackHearts = new Card("Hearts", 2, "Jack", 11);
Card jackClubs = new Card("Clubs", 3, "Jack", 11);
Card jackDiamonds = new Card("Diamonds", 4, "Jack", 11);
Card queenSpades = new Card("Spades", 1, "Queen", 12);
Card queenHearts = new Card("Hearts", 2, "Queen", 12);
Card queenClubs = new Card("Clubs", 3, "Queen", 12);
Card queenDiamonds = new Card("Diamonds", 4, "Queen", 12);
Card kingSpades = new Card("Spades", 1, "King", 13);
Card kingHearts = new Card("Hearts", 2, "King", 13);
Card kingClubs = new Card("Clubs", 3, "King", 13);
Card kingDiamonds = new Card("Diamonds", 4, "King", 13);
deck[0] = aceSpades;
deck[1] = aceHearts;
deck[2] = aceClubs;
deck[3] = aceDiamonds;
deck[4] = twoSpades;
deck[5] = twoHearts;
deck[6] = twoClubs;
deck[7] = twoDiamonds;
deck[8] = threeSpades;
deck[9] = threeHearts;
deck[10] = threeClubs;
deck[11] = threeDiamonds;
deck[12] = fourSpades;
deck[13] = fourHearts;
deck[14] = fourClubs;
deck[15] = fourDiamonds;
deck[16] = fiveSpades;
deck[17] = fiveHearts;
deck[18] = fiveClubs;
deck[19] = fiveDiamonds;
deck[20] = sixSpades;
deck[21] = sixHearts;
deck[22] = sixClubs;
deck[23] = sixDiamonds;
deck[24] = sevenSpades;
deck[25] = sevenHearts;
deck[26] = sevenClubs;
deck[27] = sevenDiamonds;
deck[28] = eightSpades;
deck[29] = eightHearts;
deck[30] = eightClubs;
deck[31] = eightDiamonds;
deck[32] = nineSpades;
deck[33] = nineHearts;
deck[34] = nineClubs;
deck[35] = nineDiamonds;
deck[36] = tenSpades;
deck[37] = tenHearts;
deck[38] = tenClubs;
deck[39] = tenDiamonds;
deck[40] = jackSpades;
deck[41] = jackHearts;
deck[42] = jackClubs;
deck[43] = jackDiamonds;
deck[44] = queenSpades;
deck[45] = queenHearts;
deck[46] = queenClubs;
deck[47] = queenDiamonds;
deck[48] = kingSpades;
deck[49] = kingHearts;
deck[50] = kingClubs;
deck[51] = kingDiamonds;

有没有更有效的方法来做到这一点

Answer 1

是：

直接分配到deck
使用一个或多个循环穿过套装＆amp;值

Answer 2

Card[] deck = new Card[52];

deck[0] = new Card("Spades", 1, "Ace", 1);

您可以为数组索引分配引用

Answer 3

以下代码可以为您服务。它只是迭代两个枚举，并为你洗牌。

使用以下抽象：

Suit
Rank
Card（Rank和Suit）
Deck：包含卡片（如果需要，随机播放）

使用简单的抽象和非复杂的代码来保存所有卡片。

public enum Rank {
    ace, two, three, four, five, six, seven, eight, nine, ten, jack, queen, king
}

public enum Suit {
    hearts, clubs, diamonds, spades
}

public class Card {
    private final Rank rank;
    private final Suit suit;

    public Card(Rank rank, Suit suit) {
        this.rank = rank;
        this.suit = suit;
    }

    public Rank rank() {
        return rank;
    }

    public Suit suit() {
        return suit;
    }
}

public class Deck {
    // Static
    private static final List<Card> allCards = new ArrayList<>();

    static {
        for (Suit s : Suit.values()) {
            for (Rank r : Rank.values()) {
                allCards.add(new Card(r, s));
            }
        }
    }

    private final List<Card> shuffledCards;

    public Deck() {
        // Shuffle
        shuffledCards= new ArrayList<>(allCards);
        Collections.shuffle(shuffledCards);
    }

    public Iterable<Card> shuffledCards() {
        return shuffledCards;
    }
}

Answer 4

为了简化你的代码，这就是我要做的事情（只是我的两分钱）：

在您的Card课程中，让您的构造函数接受两个参数，接受int并为您的suitName和faceName变量指定值，例如：

public Card(int suitValue, int faceValue) {
    switch (suitValue) {
        case 1: suitName = "Spades";   break;
        case 2: suitName = "Hearts";   break;
        case 3: suitName = "Clubs";    break;
        case 4: suitName = "Diamonds"; break;
        default: throw new RuntimeException("Invalid suit value: " + String.valueOf(suitValue));
    }
    // same for faceName
    this.suitValue = suitValue;
    this.faceValue = faceValue;
}

使用循环创建数组：

Card[] deck = new Card[52];
for (int k = 1; k <= 4; k++) {
    for (int l = 1; l <= 13; l++) {
        card[(k-1)*13+(l-1)] = new Card(k,l);
    }
}

Answer 5

我会这样做：

    String[] suitNames = {"Spades", "Hearts", "Clubs", "Diamonds" };
    String[] faceNames = {"ace", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "jack", "queen", "king"};
    List<Card> deck = new ArrayList<Card>();
    for (int i = 0; i < suitNames.length; i++) {
        String suitName = suitNames[i];
        int suitValue = i + 1;
        for (int j = 0; j < faceNames.length; j++) {
            String faceName = faceNames[j];
            int faceValue = j + 1;
            Card c = new Card(suitName, suitValue, faceName, faceValue);
            deck.add(c);
        }
    }

在java中定义大量变量

5 个答案: