java斗地主如何实现对手中的牌可有的牌型

Question

public static void main(String[] args) {
    String[] pai = new String[] { "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K", "A", "2" };
    // String[] hs = new String[] { "♥", "♠", "♦", "♣" };
    HashMap<String, Integer> hj = new HashMap<>();
    HashMap<String, Integer> hy = new HashMap<>();
    HashMap<String, Integer> hb = new HashMap<>();

    List<String> a = new ArrayList<String>();
    for (int i = 0; i < pai.length; i++) {
        a.add(pai[i]);
        a.add(pai[i]);
        a.add(pai[i]);
        a.add(pai[i]);
    }
    a.add("大王");
    a.add("小王");

    Random rand = new Random();
    String temp = null;
    for (int k = 0; k < 100; k++) {
        int p = a.size();
        int l = rand.nextInt(p);
        int m = rand.nextInt(p);
        if (l == m)
            continue;
        {
            temp = a.get(l);
            a.set(l, a.get(m));
            a.set(m, temp);
        }
    }
    // Collections.shuffle(a);

    List<String> j = new ArrayList<String>();
    List<String> y = new ArrayList<String>();
    List<String> b = new ArrayList<String>();
    List<String> d = new ArrayList<String>();
    for (int i = 0; i < a.size(); i++) {
        if (i >= a.size() - 3) {
            d.add(a.get(i));
        } else if (i % 3 == 0) {
            j.add(a.get(i));
        } else if (i % 3 == 1) {
            y.add(a.get(i));
        } else if (i % 3 == 2) {
            b.add(a.get(i));
        }
    }
    System.out.println(d);
    System.out.println("甲" + j);
    System.out.println("乙" + y);
    System.out.println("丙" + b);
    for (String str : j) {
        if (hj.containsKey(str)) {
            hj.put(str, hj.get(str) + 1);
        } else
            hj.put(str, 1);
    }
    System.out.println("j" + hj);

    for (String str : y) {
        if (hy.containsKey(str)) {
            hy.put(str, hy.get(str) + 1);
        } else
            hy.put(str, 1);
    }
    System.out.println("y" + hy);

    for (String str : b) {
        if (hb.containsKey(str)) {
            hb.put(str, hb.get(str) + 1);
        } else
            hb.put(str, 1);
    }
    System.out.println("b" + hb);

我已经实现的代码，请问用枚举法可以实现得出手中牌的牌型吗，可以的话代码是什么急求，就在我这个代码上面加！！！

Answer 1

public static void main(String[] args) {
String[] pai = new String[] { "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K", "A", "2" };
// String[] hs = new String[] { "♥", "♠", "♦", "♣" };
HashMap hj = new HashMap<>();
HashMap hy = new HashMap<>();
HashMap hb = new HashMap<>();

    List<String> a = new ArrayList<String>();
    for (int i = 0; i < pai.length; i++) {
        a.add(pai[i]);
        a.add(pai[i]);
        a.add(pai[i]);
        a.add(pai[i]);
    }
    a.add("大王");
    a.add("小王");

    Random rand = new Random();
    String temp = null;
    for (int k = 0; k < 100; k++) {
        int p = a.size();
        int l = rand.nextInt(p);
        int m = rand.nextInt(p);
        if (l == m)
            continue;
        {
            temp = a.get(l);
            a.set(l, a.get(m));
            a.set(m, temp);
        }
    }
    // Collections.shuffle(a);
    List<String> j = new ArrayList<String>();
    List<String> y = new ArrayList<String>();
    List<String> b = new ArrayList<String>();
    List<String> d = new ArrayList<String>();
    for (int i = 0; i < a.size(); i++) {
        if (i >= a.size() - 3) {
            d.add(a.get(i));
        } else if (i % 3 == 0) {
            j.add(a.get(i));
        } else if (i % 3 == 1) {
            y.add(a.get(i));
        } else if (i % 3 == 2) {
            b.add(a.get(i));
        }
    }
    Collections.sort(j);
    Collections.sort(y);
    Collections.sort(b);
    Collections.sort(d);
    System.out.println(d);
    System.out.println("甲" + j);
    System.out.println("乙" + y);
    System.out.println("丙" + b);

    for (String str : j) {
        if (hj.containsKey(str)) {
            hj.put(str, hj.get(str) + 1);
        } else
            hj.put(str, 1);
    }
    System.out.println("j" + hj);

    for (String str : y) {
        if (hy.containsKey(str)) {
            hy.put(str, hy.get(str) + 1);
        } else
            hy.put(str, 1);
    }
    System.out.println("y" + hy);

    for (String str : b) {
        if (hb.containsKey(str)) {
            hb.put(str, hb.get(str) + 1);
        } else
            hb.put(str, 1);
    }
    System.out.println("b" + hb);

    System.out.println(getDataType(j));
    System.out.println(getDataType(y));
    System.out.println(getDataType(b));
}

private static int getDataType(List<String> list){
    String[] sortlist = {"3","4","5","6","7","8","9","10","J","Q","K","A"};
    int typenum=0;
    int flag = 0;
    StringBuffer sb = new StringBuffer();
    for(int i=0;i<sortlist.length-1;){
        if(list.contains(sortlist[i]) ){
            sb.append(sortlist[i]+",");
            flag +=1;
        }else{
            if(flag>=5){
                typenum++;
                System.out.print("顺子->"+sb.toString()+"\n");
                sb.setLength(0);
            }
            flag =0;
        }
        i++;
    }
    return typenum;
}

    就加了一个统计顺子的方法 如果要算顺子 就要根据flag 值排列组合了  =5，1个 =6  3个