Question

我在改组两副牌时遇到问题。我使用二维数组创建了两个套牌。请记住，这些就是要求。我还需要使用一种称为shuffle（）的方法对它们进行混洗，该方法接受一维数组并返回一维数组。

将两个牌组混洗后，我需要从第一张牌组中选择前两张牌，并从第二张牌组中选择前两张牌，并检查它们是否匹配。以及获得结果所需的随机播放次数。

示例输出：

甲板1上的两张完全匹配的卡片是：钻石王牌，俱乐部2张

甲板2上的两张完全匹配的卡片是：钻石王牌，俱乐部2个

洗牌次数：387

这是我们正在研究的项目的第二部分，下面是我试图解决此问题的方法。

我尝试使用下面的代码创建套牌

int[][] deck = new int[2][52];

for (int i = 0; i <= deck.length - 1; i++) {
    for (int j = 0; j < deck[i].length; j++) {
        deck[i][j] = j;
    }
}

我写了shuffle方法，但是似乎不起作用。

public static int[] shuffle(int[] deck) {
  for(int i=0; i<deck.length; i++) {  
  int index = (int)(Math.random()* deck.length); 
  int temp = deck[i];  
  deck[i] = deck[index];
  deck[index] = temp;
  }
return deck;

}

下面的代码是该项目第一部分的原始代码，要求我们在将一个纸牌混排并计算获得一个四号后需要进行多少次混洗后才能从一个纸牌中打印出四种。 >

class Main {
  public static void main(String[] args) {

  String[] suit = { "Spades", "Hearts", "Diamond", "Clubs" };

  String[] rank = { "Ace", "1", "2", "3", "4", 
                 "5", "6", "7", "8", "9",
                 "10", "Jack", "Queen", "King" };

  int rank1, rank2, rank3, rank4, suit1, suit2, suit3, suit4, count = 0; 

  int[] deck = new int[52];

  for (int i = 0; i < deck.length; i++) {
     deck[i] = i;
  }

  do {

    count++;
    for (int i = 0; i < deck.length; i++) {

       int index = (int) (Math.random() * deck.length);
       int temp = deck[i];
       deck[i] = deck[index];
       deck[index] = temp;
    }

    suit1 = deck[0] / 13;
    suit2 = deck[1] / 13;
    suit3 = deck[2] / 13;
    suit4 = deck[3] / 13;

    rank1 = deck[0] % 13;
    rank2 = deck[1] % 13;
    rank3 = deck[2] % 13;
    rank4 = deck[3] % 13;

  } while (rank1 != rank2 || rank2 != rank3 || rank3 != rank4);

    System.out.println(" Four-of-kind cards: " + suit[suit1] + " of "
        + rank[rank1] + ", " + suit[suit2] + " of " + rank[rank2]
        + ", " + suit[suit3] + " of " + rank[rank3] + ", "
        + suit[suit4] + " of " + rank[rank4]
        + "\n Number of shuffled times: " + count);

    }
}

再次结果应如下所示。

示例输出：

甲板1上的两张完全匹配的卡片是：钻石王牌，俱乐部2个卡组2中有两张完全匹配的卡片：钻石王牌，俱乐部2张

洗牌次数：387

Answer 1

您的shuffle方法正确。无需执行其他将接受2D数组的随机播放方法。您只需要在2D数组的每个平台中调用shuffle方法即可。只需执行以下操作即可：

deck[0] = shuffle(deck[0]);
deck[1] = shuffle(deck[1]);

我还更新了您的代码，以便在出现其他要求时可以更轻松地进行修改。请参考下面的代码：

public class Main {
    public static String[] card; // The holder for all cards

    public static void main(String[] args) {
        card = generateCardSuits(); // generate the cards with their corresponding suits
        doPartOne();
        doPartTwo();
    }

    /**
     * Part One
     *
     */
    public static void doPartOne() {
        System.out.println("========== PART ONE ==========");
        int[] deck = generateAndShuffleInitialDeck();
        int ctr = 1;
        while (true) {
            if (deck[0] % 13 == deck[1] % 13 && deck[1] % 13 == deck[2] % 13 && deck[2] % 13 == deck[3] % 13) {
                break;
            }
            deck = shuffle(deck);
            ctr++;
        }
        System.out.println("Four-of-kind cards are: " + card[deck[0]] + " , " + card[deck[1]] + " , " + card[deck[2]]
                + " and " + card[deck[3]]);
        System.out.println("Number of shuffled times: " + ctr);
        System.out.println("==============================");
    }

    /**
     * Part Two
     *
     */
    public static void doPartTwo() {
        System.out.println("========== PART TWO ==========");
        int[][] deck = new int[2][52];
        deck[0] = generateAndShuffleInitialDeck();
        deck[1] = generateAndShuffleInitialDeck();

        int ctr = 1;
        while (deck[0][0] != deck[1][0] || deck[0][1] != deck[1][1]) {
            deck[0] = shuffle(deck[0]);
            deck[1] = shuffle(deck[1]);
            ctr++;
        }

        System.out.println("Two exact match cards from deck 1 are: " + card[deck[0][0]] + " and " + card[deck[0][1]]);
        System.out.println("Number of shuffled times: " + ctr);
        System.out.println("==============================");
    }

    /**
     * Generate an initial deck of cards and shuffle them
     *
     * @return The initial and shuffled deck of cards
     */
    public static int[] generateAndShuffleInitialDeck() {
        int[] deck = new int[52];

        for (int j = 0; j < deck.length; j++) {
            deck[j] = j;
        }
        return shuffle(deck);
    }

    /**
     * Generate the cards with their corresponding suits
     *
     * @return The deck that will serve as the mapper for the cards to their suits
     */
    public static String[] generateCardSuits() {
        String[] rank = { "Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10", "Jack", "Queen", "King" };
        String[] suit = { "Spades", "Hearts", "Diamond", "Clubs" };

        String[] cards = new String[52];

        for (int i = 0, j = 0, s = 0; i < cards.length; i++) {
            cards[i] = rank[j] + " of " + suit[s];
            j++;
            if (j == 13) { // Since each suit consist of 13 cards
                j = 0;
                s++;
            }
        }
        return cards;
    }

    /**
     * Shuffle the deck of cards
     *
     * @param deck
     *            the deck of cards to be shuffle
     *
     * @return The shuffled deck of cards
     */
    public static int[] shuffle(int[] deck) {
        for (int i = 0; i < deck.length; i++) {
            int index = (int) (Math.random() * deck.length);
            int temp = deck[i];
            deck[i] = deck[index];
            deck[index] = temp;
        }
        return deck;
    }
}

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1 个答案: