我正在尝试制作一款GoFish游戏。我正在制作套牌。我知道如何*把所有数字*＆＃34;卡＆＃34;在ArrayList中。但由于有四种不同类型的牌不是整数（King，Queen，Jack，Ace），这会导致一些速度障碍。
将这四张卡添加到此ArrayList的最佳方法是什么？
ArrayList<Integer> cards = null;
另外，我如何确保套牌中每张卡有四张？
@ user3189142如何从计算机手中搜索并取出猜测的卡片并将其添加到玩家的手中？
public static void playOneGame(ArrayList<Integer> cards, Scanner input)
{
ArrayList<Integer> computer = new ArrayList<Integer>();
ArrayList<Integer> person = new ArrayList<Integer>();
ArrayList<Integer> computerPile = new ArrayList<Integer> ();
ArrayList<Integer> personPile = new ArrayList<Integer>();
// TODO: Deal Cards
for (int p = 0; p < 7; p++)
{
person.add(cards.get(0));
cards.remove(0);
}
for (int c = 0; c < 7; c++)
{
computer.add(cards.get(0));
cards.remove(0);
}
// TODO: Show the person their starting hand
System.out.println("Your cards are: " + person);
// next line is for testing, not apart of final program
System.out.println("Computer's cards are: " + computer);
//TODO: Play one turn with the person doing the choosing
//TODO: Let the player draw from the deck
while (computerPile.size() + personPile.size() < 52 || !cards.isEmpty())
{
if (!person.isEmpty())
{
//TODO: Play one turn with the person doing the choosing
System.out.println("What card do you want?");
int card = input.nextInt();
playPersonTurn(card, person, computer, personPile, computerPile, cards);
}
else
{
//TODO: Let the player draw from the deck
person.add(cards.get(0));
cards.remove(0);
}
showGameState(person, computerPile, personPile);
//TODO: Play one turn with the computer doing the choosing
if (!computer.isEmpty())
{
int card = computer.get((int)(Math.random()*computer.size()));
System.out.println("Do you have any " + card + "'s ?");
playComputerTurn(card, person, computer, personPile, computerPile, cards);
}
else if (!cards.isEmpty())
{
//TODO: Let the computer draw from the deck
computer.add(cards.get(0));
cards.remove(0);
}
//TODO: Create piles from hands if applicable
checkPiles(person, personPile, computer, computerPiles);
showGameState(person, computerPile, personPile);
}
桩检查方法：
public static void checkPiles(ArrayList<Integer> person, ArrayList<Integer> personPile, ArrayList<Integer> computer, ArrayList<Integer> computerPile)
{
for (int i = 1; i < 14; i++)
{
if (Collections.frequency(person,i) == 4)
{
for (int j = 0; j < 4; j++)
{
personPile.add(person.remove(person.indexOf(i)));
}
else if (Collections.frequency(computer,i) == 4)
{
for (int k = 0; k < 4; k++)
{
computerPile.add(computer.remove(computer.indexOf(i)));
}
}
}
}
现在代码运作良好，除非形成一堆，它显示所有四个相同数量的卡，如下面显示的数字1.有没有办法，我们只能显示一个1？< / p>
What card do you want?
1
You took the computer's 1.
Here are your cards:
2 1 13 12 12 1 6 1 6 10 9 1
You don't have any piles yet. Computer doesn't have any piles yet.
Do you have any 7's?
Computer went fishing. Computer drew one card from deck.
Here are your cards:
2 13 12 12 6 6 10 9
Here is your pile:
1 1 1 1
Computer doesn't have any piles yet.
What card do you want?
答案 0 :(得分：1)
如果我这样做，我会将Ace添加为1，将Jack添加为11，将Queen添加为12，将King添加为13。确保套牌中每张卡有4个非常简单，使用嵌套循环：
for (int i = 0; i < 4; i++)
{
for (int j = 1; j < 14; j++)
cards.add(j);
}
这会将13张牌加入牌组4次。我假设你也需要输出卡片，这就是你遇到麻烦的原因。在输出时，您可以使用一些简单的if来检查Ace，King，Queen和Jack卡，并输出这些名称：
outputCard(int cardNumber)
{
if (cardNumber == 1)
System.out.println("Ace");
else if (cardNumber == 11)
System.out.println("Jack");
else if (cardNumber == 12)
System.out.println("Queen");
else if (cardNumber == 13)
System.out.println("King");
else
System.out.println(cardNumber);
}
要将前七张牌添加到另一个阵列列表，我们将其称为hand，请尝试以下操作：
for (int k = 0; k < 7; k++)
hand.add(cards.remove(0));
然后hand应该在cards列表
要向计算机索取一张卡，如果计算机有卡，请将其放入播放器的手中，请尝试以下操作：
//Get the card choice from the user
int cardChoice = input.nextInt();
if (computer.contains(cardChoice))
player.add(computer.remove(computer.indexOf(cardChoice)));
这样做首先检查计算机手中至少有1张你要求的牌，如果有，它会找到该牌的索引（computer.indexOf），然后将其删除（{{ 1}}），返回存储在该位置的整数，并将其添加到玩家手computer.remove。如果您需要多行，请告诉我，我会稍微改变一下。
好的，为了检查一只手是否有4张相同的牌，我们会使用一对圈来进行循环，如下所示：
player.add对于计算机手，只需复制粘贴并将//For the players hand
for (int i = 1; i < 14; i++)
{
if (Collections.frequency(player,i) == 4)
{
for (int j = 0; j < 4; j++)
{
playerPile.add(player.remove(i))
}
}
}
更改为playerPile，将computerPile更改为player
您获得的错误可能是因为您的computer函数具有返回值，但该值未分配给任何内容。它也可能不需要是2个函数，你应该能够将它们合并在一起：
pileCheck通过将循环更改为：
，您可以使堆中仅显示其中一种类型的卡public static void checkPiles()
{
for (int i = 1; i < 14; i++)
{
if (Collections.frequency(player,i) == 4)
{
for (int j = 0; j < 4; j++)
{
playerPile.add(player.remove(player.indexOf(i))) //Updated this
}
}
else if (Collections.frequency(computer,i) == 4)
{
for (int k = 0; k < 4; k++)
{
computerPile.add(computer.remove(computer.indexOf(i))) //Updated this
}
}
}
}
请注意，如果您这样做，则while while循环将需要更改为：
if (Collections.frequency(player,i) == 4)
{
for (int j = 0; j < 4; j++)
{
player.remove(player.indexOf(i));
}
playerPile.add(i);
}
因为当找到所有匹配时，两个桩中的总数只会增加到13而不是52
答案 1 :(得分：0)
使用字符串的ArrayList而不是Integers。卡片组由52张扑克牌组成，每张扑克牌都有等级和套装。要编写createDeck（）方法，请参阅下面的伪代码：
ArrayList<String> cardDeck;
enum Rank
{
ACE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING;
}
enum Suit
{
CLUB , DIAMOND , HEART , SPADE;
}
createDeck()
{
...... Write your logic to create a slide deck here ...........
for(Suit aSuit : Suit.values())
for (Rank aRank : Rank.values())
cardDeck.add(aRank + somedelimeter + aSuit)
}