首次尝试玩Blackjack游戏

Question

我对这段代码不满意，因为我确信有更好的方法来实现我想要实现的目标。我是一个初学者，我已经用我所知道的日期来完成这个任务。

import java.util.Random;
import java.util.Scanner;


public class SeventySix {

    public static void main(String[] args) {

        Scanner keyboard = new Scanner(System.in);

        //52 Cards, Aces = 11, Picture cards = 10, Ace's cannot be reduced to 1.
        int[] newCard = {2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5,6,6,6,6,7,7,7,7,8,8,8,8,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,11,11,11};

        //Shuffle. Once per game.
        shuffleArray(newCard);

        //Start BlackJack.
        System.out.println("Welcome to BlackJack!");
        System.out.println();
        System.out.println("You get a " + newCard[0] + " and a " + newCard[1] + ".");
        int playerTotal = newCard[0] + newCard[1];
        System.out.println("Your total is " + playerTotal + ".");
        System.out.println();

        //Player can get blackjack/bust in the 1st deal. - awaiting betting system (enhanced bets for blackjack in first round)
        if (playerTotal == 21){
            System.out.println("Blackjack, you win.");
            System.exit(0);
        }
        if (playerTotal > 21){
            System.out.println("Bust, You lose.");
            System.exit(0);
        }
        // Dealer cards
        System.out.println("The dealer has a " + newCard[2] + " showing, and a hidden card.");
        int dealerTotal = newCard[2] + newCard[3];
        if (dealerTotal > 21){     //Dealer bust check.
            System.out.println();
            System.out.println("Dealers total is " + dealerTotal + ".");
            System.out.println("Dealer is bust, you win!");
            System.exit(0);
        }
        if (dealerTotal == 21){    //Dealer blackjack check.
            System.out.println();
            System.out.println("Dealer reveals his second card: " + newCard[3] + ".");
            System.out.println("Dealers total is " + dealerTotal + ".");
            System.out.println();
            System.out.println("Dealer has BlackJack, you lose.");
            System.exit(0);
        }
        System.out.println("His total is hidden.");
        System.out.println();


        // Hit or Stay for player.
        System.out.print("Would you like to \"hit\" or \"stay\"? ");
        String hitStay = keyboard.next();
        System.out.println();

        //cc = card count
        int cc = 4; 
        if (hitStay.equalsIgnoreCase("hit")){
            // While loop to ensure different cards & multiple "hits".
            while (playerTotal < 21 && hitStay.equalsIgnoreCase("hit")){
                if (hitStay.equalsIgnoreCase("hit")){
                    System.out.println("You drew a " + newCard[cc] + ".");
                    playerTotal = playerTotal + newCard[cc];
                    System.out.println("Your total is " + playerTotal + ".");
                    System.out.println();
                    cc++;        //Adds 1 to ensure next card is different.
                    // Bust & Blackjack check.
                    if (playerTotal > 21){
                        System.out.println("You are bust, You lose.");
                        System.exit(0);
                    }
                    if (playerTotal == 21){
                        System.out.println("Blackjack, you win.");
                        System.exit(0);
                    }
                    System.out.print("Would you like to \"hit\" or \"stay\"? ");
                    hitStay = keyboard.next();
                    System.out.println();
                }
            }        
        }

        // Dealers turn, only if Round 1 didn't end in bust/blackjack.
        keyboard.close();
        System.out.println("Ok dealers turn.");
        System.out.println("His hidden card was a " + newCard[3] + "."); // reveal hidden from round one.

        cc++; // Pretty sure its not needed.
        while (dealerTotal < 16){ // Dealer will stay on 16+ and hit if below.
            System.out.println();
            System.out.println("Dealer chooses to hit.");
            System.out.println("He draws a " + newCard[cc] + ".");
            cc++;
            dealerTotal = dealerTotal + newCard[cc];
            System.out.println();
            System.out.println("His total is " + dealerTotal);
            // bust check - no need for blackjack check due to final win sequence
            if (dealerTotal > 21){
                System.out.println();
                System.out.println("Dealer is bust, YOU WIN!");
                System.exit(0);
            }
            // stay condition.
            if (dealerTotal < 21 && dealerTotal > 16){
                System.out.println();
                System.out.println("Dealer Stays.");
            }
        }

        // final win sequence.
        System.out.println();
        System.out.println("Dealer total is " + dealerTotal);
        System.out.println("Your total is " + playerTotal);
        System.out.println();

        if (dealerTotal > playerTotal){
            System.out.println("Dealer wins.");
        } 
        if (dealerTotal == playerTotal){
            System.out.println("You both draw.");
        }
        if (dealerTotal < playerTotal){
            System.out.println("You win.");
        }
    }

    static void shuffleArray(int[] deckCards){

        /**
         * This code is obtained from the internet and is not my own though process
         * I need to understand it before I will be happy using it.
         * I'll have a try at explaining this, please correct me if I suckarino.
         * deckCards is a placeholder for the array I chose to use.
         * i = the length of the array which is 52, -1 would be 51.
         * i will be greater than 0 until the i-- completely loops it down to 0.
         * index is a random number between 1 & 52.
         * a is a random number in the array.
         * deckCards[i] is replaced with a; 
         * essentially shuffling 1 card in the array, this happens 51 times?
         */

        Random rnd = new Random();
        for (int i = deckCards.length - 1; i > 0; i--)
        {
            int index = rnd.nextInt(i + 1);
            // Swap
            int a = deckCards[index];
            deckCards[index] = deckCards[i];
            deckCards[i] = a;
        }
    }
}

我试着让这些评论在某种程度上可以理解。

Answer 1

您应该使整个应用程序更加面向对象。一个很好的起点是将一副牌和相关功能移到其他类中。

下面是从StackOverflow问题中窃取的一个甲板的实现。注意，还有一个Card对象。

public class DeckOfCards {
  private Card cards[];

  public DeckOfCards() {
    this.cards = new Card[52];
    for (int i = 0; i < 52; i++) {
        Card card = new Card(...); //Instantiate a Card
        this.cards[i] = card; //Adding card to the Deck
    }
  }
}

还可以考虑使用enums作为套装和跟踪卡值的方法。搜索谷歌的Cark排名模式将有助于这一点。

您有很多多行打印语句，这些语句可能都可以结合在一起；

System.out.println();
System.out.println("Dealer total is " + dealerTotal);
System.out.println("Your total is " + playerTotal);
System.out.println();

可能是；

System.out.println(
  String.format("\nDealer total is %s \nYour total is %s \n", dealerTotal, playerTotal)
);

为了节省复杂性，这些人应该是else ifs；

if (dealerTotal > playerTotal){
    System.out.println("Dealer wins.");
} 
else if (dealerTotal == playerTotal){
    System.out.println("You both draw.");
}
else {
    System.out.println("You win.");
}

最后，您的代码不是很干的。您需要在不同的方法中使用更多的功能，以防止重复使用。一个很好的例子是，您有多个win序列。您的赢/输条件应该会导致调用一个win/lose方法，该方法执行适当的序列。

此外，Blackjack是一个重复的游戏，第二轮与第一轮相同，您的代码只是第二次重复相同的操作。同样，应该将其移到方法中。

Answer 2

我只想指出，虽然卡片将更好地表示为对象而不是ints，而且程序组织和流程有点混乱，但是您对洗牌算法的选择是很好的。

费舍-耶茨洗牌是一种简单而不偏不倚的方法，可以将数组元素的顺序随机化，并且您有一个教科书上的实现。即使你没有自己写代码，我认为你应该因为选择正确的代码而获得荣誉。

编辑:因为您在shuffleArray()…中询问了评论的有效性

这就是你所写的，忽略了最初的噪音：

• deckCards是我选择使用的数组的占位符。
• i =数组的长度，为52，-1将为51。
• i将大于0，直到i--将其完全循环到0。
• index是介于1和52之间的随机数。
• a是数组中的一个随机数。
• deckCards[i]代之以a；

基本上洗牌1卡在数组中，这会发生51次吗？

你的词汇表略有偏离，关于i下降到0，而index在1到52之间的评论是不准确的。

• deckCards是对输入数组的引用。此函数调整其元素的位置。
• i最初是数组最后一个元素的索引。
• i迭代到1。
• index是介于0和i之间的一个随机数，包含在内。
• a是数组的随机元素。
• deckCards[i]与a交换；

基本上洗牌1卡在数组中，这会发生51次吗？

然而，这是一个缺乏洞察力的，机械的代码描述。更好的意见是：

对于数组中的每个元素(第一个除外)，从末尾开始，以一致的概率选择任何先前的元素(或其本身)，并交换它们。这种技术是无偏的。例如，考虑一下第一个循环在洗牌52张牌时会发生什么.甲板上的每一张牌都同样有可能占据deckCards[51]的位置。对于最后一个元素deckCards[51]，重复这个过程--从剩下的51张卡牌中随机抽取，来填充它的最后一个元素deckCards[50]，等等。

Answer 3

代码中有一个小错误：

Aces只有11，但不是1。这意味着玩家(或经销商)可能只需要两张牌就会崩溃。

假设他们只有Aces，那么这将是一个半身像。我会加上一张支票，看看这名球员是否正在崩溃，但其中一张牌是作为一张Ace牌发行的。甚至可能创建一个布尔值，PlayerHasAce。

伪码：

If they bust, and PlayerHasAce is TRUE;
    minus 10 from the total;
    are they still busting without additional aces? 
    No? Good! They can continue playing. 

不过，您只需要确保每个Ace只设置一次布尔值，而不是在同一只手中重复。