如何交错两个数组列表？

我正在尝试开发一个程序，通过将一副牌分成两部分然后将它们交错来洗牌。

班级甲板代表一副 52 张牌。有两种方法：牌组(int n) and 抽卡卡().

牌组(int n)是构造函数。该参数告诉我们应该洗牌多少轮。在每轮洗牌中，整副牌首先被分成两个子牌组。然后，子甲板交错成一个完整的甲板。

一些笔记:

• 为了简化讨论，我们假设卡片是 1、2、…、10。

• 第一轮，整副牌分为[1,2,3,4,5]和[6,7,8,9,10]。然后，我们通过将两个子套牌交错为 [1, 6, 2, 7, 3, 8, 4, 9, 5, 10] 来组合它们。

• 在第二轮中，我们再次将整个牌组分为两个子牌组 [1, 6, 2, 7, 3] 和 [8, 4, 9, 5, 10]，然后将它们组合为 [1, 8, 6] 、4、2、9、7、5、3、10]。

• 由于我们总是将第一副牌放在第二副牌之前，因此无论我们洗牌多少轮，第一副牌和最后一张牌都保持不变。

• 牌组的原始顺序是 S2、S3、…、SK、SA、H2、…、HA、C2、…、CA、D2、…、DA。

抽卡卡()取出牌组中的第一张牌并将其归还。参考上面讨论的第二轮后的牌组，抽卡()返回 1，牌组变为 [8, 6, 4, 2, 9, 7, 5, 3, 10]。

我的交错方法：创建 3 个数组列表，其中 2 个 (cards1 and cards2）持有 SA - HA 和 C2 - DA 牌，另一张（shuffled）持有交错的甲板。我设法实现了原始的牌组顺序，但是当我尝试隔行扫描时，出现越界错误：“索引 0 超出长度 0 的范围”。

Question： 我究竟做错了什么？

这是我的代码:

import java.util.*;

public class Deck {
    private int rounds;
    private ArrayList<Card> cards = new ArrayList<Card>();
    private ArrayList<Card> cards1 = new ArrayList<Card>();
    private ArrayList<Card> cards2 = new ArrayList<Card>();
    private ArrayList<Card> shuffled = new ArrayList<Card>();

    public Deck(int n) {
        for (Suit s : Suit.values()) {
            for (Rank r : Rank.values()) {
                cards.add(new Card(r,s));
            }
        }

        for (int x=0; x<n; x++) {
            for (int i=0; i<((cards.size())/2); i++) {
                cards1.add(cards.get(i));
                for (int j=26; j<cards.size(); j++) {
                    cards2.add(cards.get(j));
                    for (int k=0; k<cards.size(); k++) {
                        shuffled.add(k*2, cards1.get(i));
                        shuffled.add(k*2+1, cards2.get(j));
                    }
                }
            }
        }

        System.out.println(cards);
        System.out.println(cards1);
        System.out.println(cards2);
        System.out.println(shuffled);
        rounds = n;
    }

    public Card drawCard() {
        Card removed = shuffled.get(0);
        shuffled.remove(0);
        return removed;
    }
}

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

    private String suit;

    Suit (String s) {
      suit = s;
    }

    public String toString() {
      return suit;
    }
}

// YOU CANNOT MODIFY THIS FILE

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

    Deck deck; 

    deck = new Deck(0);
    System.out.println("The original deck is: ");
    for (int i = 0; i < 52; i++) {
      System.out.print(deck.drawCard() + " ");
    }
    System.out.println();
    System.out.println();

    deck = new Deck(1);
    System.out.println("After shuffling once is: ");
    for (int i = 0; i < 52; i++) {
      System.out.print(deck.drawCard() + " ");
    }
    System.out.println();
    System.out.println();

    deck = new Deck(2);
    System.out.println("After shuffling twice is: ");
    for (int i = 0; i < 52; i++) {
      System.out.print(deck.drawCard() + " ");
    }
    System.out.println();
    System.out.println();
  }
}

假设的输出为测试平台类 is

The original deck is:
S2 S3 S4 ... DK DA

After shuffling once is:
S2 C2 S3 C3 ... DA

After shuffling twice is:
S2 H2 C2 D2  ... DA

好吧，亲爱的，实际上你得到了“索引越界”（上帝知道为什么......:)，这是我解决它的方法（带注释）：

public class Deck {

    //constants for 52 and 26 :
    private static final int FULL_DECK = Suit.values().length * Rank.values().length;
    private static final int HALF_DECK = FULL_DECK / 2;
    // use the constants, we need only one list (+2 temp lists, throw away
    // "shuffeld" (not needed, confusing, we use "cards" for "full deck")):
    private final ArrayList<Card> cards = new ArrayList<>(FULL_DECK);
    public Deck(int n) {

        init(); // as you had/see below

        // more overview/structure ... and we can limit n:
        for (int rounds = 0; rounds < n % 8; rounds++) {
            interlace();
        }
        // comment this, since we do output in main method...
        // System.out.println(cards);
    }

初始化和“交错”方法：

    private void init() {
        for (Suit s : Suit.values()) {
            for (Rank r : Rank.values()) {
                cards.add(new Card(r, s));
            }
        }
    }

    private void interlace() {
        // throw exception, when illegal state
        assert (!cards.isEmpty());
        // left & right temp lists:
        final ArrayList<Card> left = new ArrayList<>(HALF_DECK);
        final ArrayList<Card> right = new ArrayList<>(HALF_DECK);
        // put the first half of "cards" into "left"
        left.addAll(cards.subList(0, HALF_DECK));
        // ...the rest into "right"
        right.addAll(cards.subList(HALF_DECK, FULL_DECK));
        // clear "cards"
        cards.clear();

        // iterate half deck:
        for (int i = 0; i < HALF_DECK; i++) {
            // fill cards from "left" (with "double step")
            cards.add(i * 2, left.get(i));
            // ..and from "right" (with "double step" +1;)
            cards.add(i * 2 + 1, right.get(i));
        }
        // done!
        // debug:
        // System.out.println(left);
        // System.out.println(right);
        // System.out.println(cards);
    }

“draw”方法将如下所示：

    public Card drawCard() {
        assert (!cards.isEmpty());
        return cards.remove(0);
    }

使用相同的主要方法（Suit、Rank 类），我们得到：

The original 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 

After shuffling once is: 
S2 C2 S3 C3 S4 C4 S5 C5 S6 C6 S7 C7 S8 C8 S9 C9 S10 C10 SJ CJ SQ CQ SK CK SA CA H2 D2 H3 D3 H4 D4 H5 D5 H6 D6 H7 D7 H8 D8 H9 D9 H10 D10 HJ DJ HQ DQ HK DK HA DA 

After shuffling twice is: 
S2 H2 C2 D2 S3 H3 C3 D3 S4 H4 C4 D4 S5 H5 C5 D5 S6 H6 C6 D6 S7 H7 C7 D7 S8 H8 C8 D8 S9 H9 C9 D9 S10 H10 C10 D10 SJ HJ CJ DJ SQ HQ CQ DQ SK HK CK DK SA HA CA DA 

它不是“那个”线程安全......但出于演示目的......希望它有帮助！ :)

..实际上索引超出范围，因为你never filled shuffeled, when n == 0... iobex 在Main: System.out.print(deck.drawCard() + " "); (and (n == 0))