Java-如何以升序向数组列表添加元素并计算数组列表中的对数？

Question

与我之前的帖子相似，我试图开发一个纸牌游戏：

a）打印出整副牌

接下来，私有静态int numPairs（ArrayList cards）是一个计算卡片中卡片对数量的方法。如果两张牌等级相同，则两张牌成对。牌中三张相同等级的牌计为1对。如果相同等级的四张牌都在牌中，则有2对牌。这里的问题是，我不确定如何用我编写的代码将3张等级相同的牌计数为1对，并且仍然会出现表示字段card.rank不可见的错误，因为我使用了。ordinal（）方法来比较牌的等级。

我如何修复这些错误和我的代码为这两个特定的方法？

import java.util.ArrayList;

public class SimpleCardGame2 {
  private static void addCard(ArrayList<Card> cards, Card c) {
        for (int x=0; x<10; x++) {
            if (cards.get(x) < c.rank.ordinal()) {
                continue;
            }
            if (cards.get(x) == c.rank.ordinal()) {
                return;
            }
            cards.add(x,c);
            return;
        }
        cards.add(c);
   }

  private static int numPairs(ArrayList<Card> cards) {
        int count=0;
        for (int j=0; j<10; j++) {
            for (int k=j+1; k<10; k++) {
                if (cards.get(j).rank.ordinal() == cards.get(k).rank.ordinal()) {
                    count+=1;
                }
            }
        }
        return count;
   }

  public static void main(String[] args) {

    ArrayList<Card> deck;  // an arraylist of Cards, representing a deck of cards

    // put a copy of each card into cards
    // The order of the cards in the array is: 
    // S2, S3, ..., SK, H2, ..., HK, C2, ..., CK, D2, ..., DK

    deck = new ArrayList<Card>();
    for (Suit s : Suit.values()) {
        for (Rank r : Rank.values()) {
            deck.add(new Card(r,s));
        }
    }

    //
    // YOU DO NOT HAVE TO MODIFY ANYTHING BELOW
    //

    // print out the whole deck once
    System.out.println("The whole deck is:");
    System.out.println(deck);

    // randomly draw 10 cards
    // cards drawn will be removed from deck 
    MyRandom rnd = new MyRandom();
    ArrayList<Card> cards = new ArrayList<Card>();
    for (int i = 0; i < 10; i++)
      addCard(cards, deck.remove(rnd.nextInt(deck.size())-1));


    // print out the card drawn and the number of pairs
    System.out.print("The cards drawn are: ");
    System.out.println(cards);
    System.out.print("There are ");
    System.out.print(numPairs(cards));
    System.out.println(" pair(s) of cards.");
  }
}

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

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

    public String toString() {
        return suit + "" + rank;
    }

    public int compareTo(Card c) {
        int diffRank = rank.ordinal() - c.rank.ordinal();
        if (diffRank < 0) {
            return -1;
        }
        else if (diffRank > 0) {
            return 1;
        }
        else if (diffRank == 0) {
            int diffSuit = suit.ordinal() - c.suit.ordinal();
            if (diffSuit < 0) {
                return -1;
            }
            else if (diffSuit > 0) {
                return 1;
            } 
        }
        return 0;
    }
}

public enum Rank {
  TWO("2"), 
  THREE("3"), 
  FOUR("4"), 
  FIVE("5"), 
  SIX("6"), 
  SEVEN("7"), 
  EIGHT("8"),
  NINE("9"), 
  TEN("10"), 
  JACK("J"), 
  QUEEN("Q"), 
  KING("K"),
  ACE("A"); 

  private String rank;

  // Constructor
  Rank (String r) {
    rank = r;
  }

  public String toString() {
    return rank;
  }
}

public enum Suit {
    SPADE("S"), 
    HEART("H"), 
    CLUB("C"), 
    DIAMOND("D"); 

    private String suit;

    Suit (String s) {
      suit = s;
    }

    public String toString() {
      return suit;
    }
}

import java.util.Random;

public class MyRandom {

  Random rnd;

  public MyRandom() {
//    rnd = new Random(180);
    rnd = new Random();
  }

  // generate a random integer from 1 to max
  public int nextInt(int max) {
    return rnd.nextInt(max) + 1;
  }
}

以下是正确程序的假定输出：

The whole deck is:
[S2, S3, S4, S5, S6, S7, S8, S9, S10, SJ, SQ, SK, SA, H2, H3, H4, H5, H6, H7, H8, H9, H10, HJ, HQ, HK, HA, C2, C3, C4, C5, C6, C7, C8, C9, C10, CJ, CQ, CK, CA, D2, D3, D4, D5, D6, D7, D8, D9, D10, DJ, DQ, DK, DA]
The drawn cards are: [D4, D5, S9, C10, DQ, CQ, HQ, SQ, HK, SA]
There are 2 pair(s) of cards.

The whole deck is:
[S2, S3, S4, S5, S6, S7, S8, S9, S10, SJ, SQ, SK, SA, H2, H3, H4, H5, H6, H7, H8, H9, H10, HJ, HQ, HK, HA, C2, C3, C4, C5, C6, C7, C8, C9, C10, CJ, CQ, CK, CA, D2, D3, D4, D5, D6, D7, D8, D9, D10, DJ, DQ, DK, DA]
The drawn cards are: [D2, D5, H9, S9, H10, HJ, DK, HK, SK, CA]
There are 2 pair(s) of cards.

Answer 1

我将为每个序列创建一个ArrayList。然后遍历抽到的牌，并将每张牌复制到正确的秩ArrayList。然后您需要检查每个rank ArrayList的大小：如果大小为2或更大，则包含一对。计算大小为2或更大的ArrayList的数量，您就知道对的数量。