Question

我试图制作一个用户有3次机会猜测程序生成的随机数的游戏。到目前为止我有这个代码，但我不知道如何在用户输入3个猜测中的3个后停止程序。如果用户无法在3次尝试中正确猜测，我希望程序说＆＃34;你松了。这个数字是......＆＃34;

import java.util.Random;
import java.util.Scanner;
class GuessNumber {
    public static void main(String args[]) {
        Random random = new Random();
        Scanner input = new Scanner(System.in);
        int MIN = 1;
        int MAX = 10;
        int comp = random.nextInt(MAX - MIN + 1) + MIN;
        int user;


        do {
            System.out.print("Guess a number between 1 and 10: ");
            user = input.nextInt();

            if (user > comp)
                System.out.println("My number is less than " + user + ".");
            else if (user < comp)
                System.out.println("My number is greater than " + user + ".");
            else
                System.out.println("Correct! " + comp + " was my number! " );
        } while (user != comp);
    }
}

Answer 1

只需计算尝试次数，并在达到阈值后退出循环。像那样：

    int attemptsNum = 0;
    final int maxAttempts = 3;
    do {
        System.out.print("Guess a number between 1 and 10: ");
        user = input.nextInt();

        if (user > comp)
            System.out.println("My number is less than " + user + ".");
        else if (user < comp)
            System.out.println("My number is greater than " + user + ".");
        else
            System.out.println("Correct! " + comp + " was my number! ");
    } while (user != comp && ++attemptsNum <maxAttempts );

    if (attemptsNum == maxAttempts)
        System.out.println("You loose. The number was :" + comp);

将猜数游戏限制为三次尝试

1 个答案: