将范围保持在一定的数字范围内

Question

对于我的Java类，我正在编写一个小程序，首先选择一个介于1和100之间的int数字。然后，它会提示用户开始猜测正确的int。如果用户对int的猜测过高或过低，程序会打印出一个新范围，供他们在该范围内进行猜测。如果用户输入字符串或双精度，程序只需重新要求用户输入整数，但不会以任何方式更改范围。

示例输出（当机密号为20时）如下所示：

c:\csc116> java GuessingGame 
Guess the secret number!
Enter a number between 1 and 100 (inclusive): 45
Enter a number between 1 and 44 (inclusive): jlkj
Enter a number between 1 and 44 (inclusive): 31.0  //double
Enter a number between 1 and 44 (inclusive): 1000 //outside the range of 1-100
Enter a number between 1 and 44 (inclusive): 34
Enter a number between 1 and 33 (inclusive): 15
Enter a number between 16 and 33 (inclusive): 20    
You win!

该项目似乎已经基本完成，但只有一个例外。其中一个要求是，当用户键入的int超出我们给定的1和100范围时，打印输出消息不会改变（如上面的示例所示）。这就是我被难倒的地方，我想看看是否有人能帮我找到正确的答案。

import java.util.*;


public class GuessingGame {


    public static void main(String[] args) {
        introduction();
        Scanner console = new Scanner(System.in);
        Random rand = new Random();

        int guess = 0;
        int minimum = 1;
        int maximum = 100;
        int secretNumber = rand.nextInt(100) + 1;

        System.out.print("Enter a number between " + minimum + " and " + maximum + " (inclusive): ");
        while (guess != secretNumber) {
            if (console.hasNextInt()) {
                guess = console.nextInt();
                if (guess > secretNumber) {
                    maximum = guess - 1;
                    System.out.print("Enter a number between " + minimum + " and " + maximum + " (inclusive): ");
                }
                if (guess < secretNumber) {
                    minimum =guess + 1;
                    System.out.print("Enter a number between " + minimum + " and " + maximum + " (inclusive): ");
                }
                if (guess == secretNumber) {
                    System.out.println("You win!");
                }
            } else {
                console.next();
                System.out.print("Enter a number between " + minimum + " and " + maximum + " (inclusive): ");
            }
        }
    }


    public static void introduction() {
        System.out.println("Guess the secret number!");
    }
}

Answer 1

您只检查相对于机密号码的猜测号码。你缺少的是相对于最大值和最小值的猜测数的检查。例如：

if (guess > maximum) {
    System.out.print("Too high!");
} else if (guess < minimum) {
    System.out.print("Too low!");
}

Answer 2

当您还想提示用户初始最小值和最大值时，应将这两个值分开保存，并在循环开始时插入另一个（如果）-检查

guess = console.nextInt();
if (guess > originalMaximum) {
    System.out.print("Enter a number less then " + originalMaximum);
}

Answer 3

你有：

            guess = console.nextInt();
            if (guess > secretNumber) {
                maximum = guess - 1;
                System.out.print("Enter a number between " + minimum + " and " + maximum + " (inclusive): ");
            }
            if (guess < secretNumber) {
                minimum =guess + 1;
                System.out.print("Enter a number between " + minimum + " and " + maximum + " (inclusive): ");
            }
            if (guess == secretNumber) {
                System.out.println("You win!");
            }

现在，您根本没有检查最小/最大范围。你必须添加一个明确的检查，但需要注意的是（在这里的其他答案中遗漏了），如果输入超出范围，你必须确保不要将其作为猜测处理。您当前使用ifs而不使用elses的风格意味着您在实现它时必须小心。你有几个选择，例如：

            guess = console.nextInt();
            if (guess < minimumAllowed || guess > maximumAllowed) {
                // handle error
            } else {
                // handle valid input
                if (guess > secretNumber) {        
                    // ...
                }
                if (guess < secretNumber) {
                    // ...
                }
                if (guess == secretNumber) {
                    // ...
                }
            }

或：

            guess = console.nextInt();
            if (guess < minimumAllowed || guess > maximumAllowed) {
                // handle error
            } else if (guess > secretNumber) {        
                // ...
            } else if (guess < secretNumber) {
                // ...
            } else if (guess == secretNumber) {
                // ...
            }

或者，坚持你目前的风格，只要你不必在循环中再做任何不相关的逻辑（在你的程序中似乎就是这样）：

            guess = console.nextInt();
            if (guess < minimumAllowed || guess > maximumAllowed) {
                // handle error
                continue;
            }
            // handle valid input
            if (guess > secretNumber) {        
                // ...
            }
            if (guess < secretNumber) {
                // ...
            }
            if (guess == secretNumber) {
                // ...
            }