扑克电动汽车计算器:计算手值？

Question

我试图写一个程序，帮助用户使正确的EV播放每一只手。然而，在这一分钟，我使用卡值(即总共两张卡)作为我的决定的基础。比如9=9，10=10，j=11，q=12.我希望能够在他们的实际手中进入，例如Adks (钻石王牌，黑桃之王)。这将是更准确的，因为它将考虑到合适的价值，手等。谁能给我的意见，最好的方式纳入这一点？事先非常感谢！我的密码在下面！

package uk.ac.qub.lectures;

//importing resources (scanner)
import java.util.Scanner;

public class PokeGame {

    public static final int MIN_POSITION = 1;
    public static final int MAX_POSITION = 8;

    public static void main(String[] args) {
        // declaring user position
        int userPosition = 0;
        // setting up scanner
        Scanner scanner = new Scanner(System.in);
        // integer referring to use again or not
        int useAgain = 0;
        // boolean getting valid input for repeat
        boolean repeat = false;

        // declaring number value of each card
        int cards;

        do {

            // getting user position
            do {
                System.out.printf("Please enter position between %d and %d\n",MIN_POSITION, MAX_POSITION);
                userPosition = scanner.nextInt();
            } while ((userPosition < MIN_POSITION)  || (userPosition > MAX_POSITION));
            // getting hand hand strength
            System.out.println("Enter card value");
            cards = scanner.nextInt();

            switch (userPosition) {

            case 1:
            case 2:
                if (cards > 10) {
                    System.out.println("SHOVE");
                } else
                    System.out.println("FOLD");
                break;
            case 3:
            case 4:
            case 5:
                if (cards > 13) {
                    System.out.println("SHOVE");
                } else
                    System.out.println("FOLD");
                break;
            case 6:
            case 7:
            case 8:
                if (cards > 17) {
                    System.out.println("SHOVE");
                } else
                    System.out.println("FOLD");
                break;
            default:
                System.out.println("ENTER VALID POSITION");
            }
            do {

                System.out.println("Do you advice on another Hand?");
                System.out.println("Enter 1 for Yes, Enter 0 for No");
                useAgain = scanner.nextInt();
                if ((useAgain == 1) || (useAgain == 0)) {

                    repeat = false;
                } else {

                    System.out.println("Invalid Input, please enter 1 or 0");
                    repeat = true;
                }
            } while (repeat);
        } while (useAgain != 0);

        // clean up resources
        scanner.close();
    }// method end

}// class end

Answer 1

如果您将卡片输入如下："AA“、"9T”或“9T”，则可以使用输入卡根据合适性和空白计算手值：

import java.util.Arrays;
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
String[] hand = (scanner.nextLine() + 'u').toUpperCase().split("");
String values = "  23456789TJQKA";
int[] cards = new int[] {values.indexOf(hand[1]), values.indexOf(hand[2])};
Arrays.sort(cards);
int gap = cards[1] - cards[0] - 1;
boolean pair = gap == -1;
boolean suited = hand[3].equals("S");
char[] cards = new char[] {(char)values.charAt(cards[0]), (char)values.charAt(cards[1])};
int handValue = 0;

// adjust value based on pairs, suitedness or connectedness
if (pair) // hand is a pair
    handValue += 10; //or whatever you want
else if (suited) // hand is suited
    handValue += 3; //or whatever you want

if (gap == 0) // hand is no gap
    handValue += 5; //or whatever you want.

if (gap == 1) // hand is one gap
    handValue += 3; //or whatever you want.

if (cards[1] == 'A' && cards[0] == 'K' && suited) // AK suited
    System.out.println("AK Suited!");
else if (cards[1] == 'A' && suited) // Ax suited
    System.out.println("Ax Suited!");