Java 询问用户是否想再次播放

我写了一个扑克程序，把牌发给输入的玩家，然后发房子牌。我想知道如何在最后询问玩家是否想再次玩，并将程序放入循环中。因此，如果输入“是”，程序将重新启动，但如果输入“否”，程序将结束。我该怎么做

import java.io.*;

public class Dealer {

    public static void main(String[] args) throws IOException {

        BufferedReader in;
        int x;
        String playerx;

        in = new BufferedReader(new InputStreamReader(System.in));
        System.out
                .println("Welcome to the Casino! My name is Zack and I'm going to be dealing your table. How many players are playing?");
        playerx = in.readLine(); // user input for menu selection
        x = Integer.valueOf(playerx).intValue();

        while (x >= 1 && x <= 24) {

            // create a deck of 52 cards and suit and rank sets
            String[] suit = { "Clubs", "Diamonds", "Hearts", "Spades" };
            String[] rank = { "2", "3", "4", "5", "6", "7", "8", "9", "10",
                    "Jack", "Queen", "King", "Ace" };

            // initialize variables
            int suits = suit.length;
            int ranks = rank.length;
            int n = suits * ranks;
            // counter (5 house cards and 2 cards per player entered)
            int m = 5 + (x * 2);

            // initialize deck
            String[] deck = new String[n];
            for (int i = 0; i < ranks; i++) {
                for (int j = 0; j < suits; j++) {
                    deck[suits * i + j] = rank[i] + " of " + suit[j];

                }
            }

            // create random 5 cards
            for (int i = 0; i < m; i++) {
                int r = i + (int) (Math.random() * (n - i));
                String t = deck[r];
                deck[r] = deck[i];
                deck[i] = t;
            }

            // print results
            for (int i = 0; i < m; i++) {
                System.out.println(deck[i]);
            }
        }
    }
}

import java.io.*；
公共类交易商{
公共静态void main（字符串[]args）引发IOException{
缓冲读取器；
int x；
弦乐演奏家；
in=新的BufferedReader（新的InputStreamReader（System.in））；
系统输出
.println（“欢迎来到赌场！我的名字是扎克，我将为你的桌子做交易。有多少玩家在玩？”）；
playerx=in.readLine（）；//用于菜单选择的用户输入
x=整数.valueOf（playerx）.intValue（）；
虽然（x>=1&&x我想你以前没有编程经验吗？我建议你阅读sun文档。在你考虑添加一个再次播放的选项之前，先了解变量、方法、对象和构造函数。如果你是初学者，Youtube教程也可以帮你，但不要仅仅依赖它们。
杀死两只鸟w一石二鸟：给运行程序的人在任何牌发牌之前退出的能力，以防他们不想玩。你已经有了适当的结构

while(1) {
    System.out.println("Welcome to ... How many players are playing (1-24) or enter 0 to exit?");
    x = Integer.valueOf(playerx).intValue();
    if(x == 0 || x >= 24) {
        break;
    }
    // rest of your logic remains.....
}