Question

对于我的Java类，我正在编写一个小程序，首先选择1到100之间的int数字。然后它会提示用户开始猜测正确的int。如果用户猜测int太高或太低，程序会打印出一个新的范围供他们猜测。如果用户输入String或double，程序只会重新要求用户输入int，但不会以任何方式更改范围。

示例输出（当密码是20时）看起来像：

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!

该程序似乎几乎存在，但有一个例外。其中一个要求是，当用户键入超出我们给定范围1和100的int时，打印输出消息不会更改（如上例所示）。这是我很难得到的地方，我正在寻找是否有人可以帮助指导我找到正确的答案。

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

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!");
            }

现在你根本没有检查最小/最大范围。您将不得不为此添加一个明确的检查，但警告（在此处的其他答案中遗漏）是您必须确保不处理输入作为猜测它是否已经出来范围。您目前使用if而没有else的样式意味着在实施时必须小心。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) {
                    // ...
                }
            }

或者：

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

或者，坚持你当前的风格，只要你不必在循环中做任何更多不相关的逻辑（在程序中似乎就是这种情况）：

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

Answer 2

如果您还希望将用户提示为原始最小值和最大值，则应分别保留这两个值并插入另一个if - 在循环开始时检查

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

Answer 3

您只是检查相对于密码的猜测号码。您缺少的是检查相对于最大值和最小值的猜测数字。例如：

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

Answer 4

在您检查猜到的号码是否少于或多于秘密号码之前，请在此之前进行此检查：

if (guess < minimum || guess > maximum) 
{
    System.out.print("Please enter a number between " + minimum + " and " + maximum);
    continue;
}

if (guess > secretNumber) 
{
   maximum = guess - 1;
   System.out.print("Enter a number between " + minimum + " and " + maximum + " (inclusive): ");
}

将范围保持在一组数字内

4 个答案: