Question

我的程序应该执行以下步骤：

从0到100生成随机数。
随机显示否，并要求用户输入(h/l/c)？（用户必须输入其中一个）。
如果正确，请询问用户是否愿意再次播放(y/n)？（用户必须回答(y/n)）

我能够执行问题1。问题2，随机没有显示，但我无法输入字符(h/l/c)。另外，我不能问玩家是否想再玩一次？

这就是我所做的：

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


public class NumberGuessingGame {

    public static void main(String[] args) {
        Scanner keyboard = new Scanner(System.in);

        int number;
        int guess = 0;
        int min = 0;
        int max = 100;
        int answer= (min+max);

        number = (int) (Math.random() *100 +1);
        System.out.println("Guess a number between 1 and 100.");

        //Ask user to type higher 'h', lower 'l' and correct 'c'.
        System.out.println("Is it " + number + " ?" + " (h/l/c): " );

        //
        guess = 50;

        while (guess <= 7)
        {
            System.out.println( "It is: " + guess + "?" + "(h/l/c) : " );

            //Type 'h' if guess is high, 'l' for low and 'c' for correct. 
            if(answer == 'h')
            {
                max = guess -1;
                min = 0;
                guess = ((max = min)/2) + min;
                guess++;

            } else if (answer == 'l')
            {
                max = 100;
                min = guess + 1;
                guess = ((max+min)/2);
                guess++;
            } else if (answer == 'c');
        }
        System.out.println("Great! Do you want to play again? (y/n): ");

        if(answer == 'y')
        {
            System.out.println("Guess a number between 1 and 100.");
            //else prompt another question with if else
        } else{
             System.exit(0);
        }
    }
}

Answer 1

如果猜测太低或太高，您的程序不会接受任何用户输入。

你宣布＆＃34;回答＆＃34;等式，所以它应该接受一个数字。答案需要char或String值。

answer = keyboard.nextChar(); //read a char
answer = keyboard.nextLine(); //read a String

我希望这对你有所帮助。

我怎样才能完成猜数游戏？

1 个答案: