Question

我试图创造一副牌。卡组应该是一个数组，每张卡应该是一个对象

Card(int value, String name, int suit)

int值是卡的等级：2 = 2，Three = 3 ,. 。。，杰克= 11

String名称是“Two”，“Three，..，”Ace“。

int诉讼可以是：0,1,2或3.其中0 =俱乐部，1 =钻石，2 =心，3 =黑桃

我知道如何创建Card对象。我遇到的问题是什么时候我应该创建Card [] deck数组。显然我可以一次做一个：

Card[] deck = new Card[52];
deck[0] = new Card(2, "Two", 0); // Two of clubs, value = 2
deck[1] = new Card(3, "Three", 0); // Three of clubs, value = 3

但是用循环来做这件事会很好。但我无法做对..

Answer 1

您需要首先创建一个数组，以使用其名称映射卡的值。

E.g。

String[] cardNames = new String[] {"", "", "Two", "Three", ...};

然后你可以使用嵌套循环。

Card[] deck = new Card[52];
int count = 0;
for(int i = 0; i < 4; i++) {
  for(int j = 2; j < 15; j++) {
    deck[count] = new Card(j, cardNames[j], i);
  }
  count++;
}

Answer 2

试试这个（循环可以做得更好，但想给你一个想法），

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;

public class CardExample {

    public static void main(String[] args) {

        Card[] deck = new Card[52];
        Map<Integer, String> cardSeq = new HashMap<Integer, String>();
        cardSeq.put(1, "Ace");
        cardSeq.put(2, "Two");
        cardSeq.put(3, "Three");
        cardSeq.put(4, "Four");
        cardSeq.put(5, "Five");
        cardSeq.put(6, "Six");
        cardSeq.put(7, "Seven");
        cardSeq.put(8, "Eight");
        cardSeq.put(9, "Nine");
        cardSeq.put(10, "Ten");
        cardSeq.put(11, "Jack");
        cardSeq.put(12, "Queen");
        cardSeq.put(13, "King");

        int counter=0;
        for (int i = 0; i <= 3; i++) {
            Iterator<Map.Entry<Integer, String>> it = cardSeq.entrySet().iterator();
            while (it.hasNext()) {
                Map.Entry<Integer, String> pair = it.next();
                deck[counter]=new Card(pair.getKey(), pair.getValue(), i);
                counter++;
            }

        }

        for(Card c:deck){
            System.out.println(c);
        }
    }
}

class Card {
    private int value;
    private String name;
    private int suit;

    Card(int value, String name, int suit) {
        this.value = value;
        this.name = name;
        this.suit = suit;
    }

    @Override
    public String toString() {
        return "Card [value=" + value + ", name=" + name + ", suit=" + suit + "]";
    }

}

Answer 3

首先，实际上你没有必要明确存储排名（Ace，Two，Three ......），你可以创建一个方法，根据排名值返回排名。我会重新设计你的Card课程，就像这样。

class Card {
    private int suit, rank;

    Card(int suit, int rank) {
        this.suit = suit;
        this.rank = rank;
    }

    public String getRankName() {
        switch(rank)
        {
        case 11:
            return "J";
        case 12:
            return "Q";
        case 13:
            return "K";
        case 14:
            return "A";
        default:
            return Integer.toString(rank);
        }
    }

    //other getters, setters, methods....
}

接下来只需使用嵌套的for循环来创建Card个对象

Card[] deck = new Card[52];
int index = 0;
for(int i = 0; i < 4; ++i) {
  for(int j = 2; j < 15; ++j) {
    deck[index] = new Card(i, j);
  }
  ++index;
}

创建对象数组9

3 个答案: