java - 应用函数方法(调用函数)求阶乘

Question

public static void main (String args [])  {
    Scanner sc = new Scanner(System.in);
    int number;
    int factor=1;
    System.out.println("Enter a number to find the factorial of it: ");
    number= sc.nextInt();
    factor=factorial(number);
    if (number < 1 || number > 10)
    {
        System.out.println("Invalid!! the number has to be between 1 and 10");
    }
    System.out.println("The factorial of  "+number+" is = " +factor);        
}

public static int factorial (int number) {
    int result = 1;
    if (number < 1 || number > 10)
    {
        System.out.println("Invalid!! the number has to be between 1 and 10");
    }
    else {
        for(int x=1; x<=number; x++ ) {
            result= result*x;
        }
    }
    return result;
}

我的代码工作正常，但如果我输入 11，它会显示无效消息，并且还会计算我不想要的 11 的阶乘。

Answer 1

您应该移动检查用户输入范围的代码，使其位于对 factorial() 函数的调用之上:

public static void main (String args [])  {
    Scanner sc = new Scanner(System.in);
    int number;
    int factor=1;
    System.out.println("Enter a number to find the factorial of it: ");
    number= sc.nextInt();
    if (number < 1 || number > 10) {
        System.out.println("Invalid!! the number has to be between 1 and 10");
    } else {
        factor=factorial(number);
        System.out.println("The factorial of  "+number+" is = " +factor);
    }
}

请注意，对 factorial() 的调用和 println() 语句现在都位于新的 else block 内。因此，如果用户输入无效数字，程序给出的唯一响应就是错误消息。

如果您还想在 factorial() 中进行错误检查，最清晰的方法可能是抛出 IllegalArgumentException当给出无效输入时:

public static int factorial (int number) {
    int result = 1;
    if (number < 1 || number > 10)
    {
        System.out.println("Invalid!! the number has to be between 1 and 10");
        throw new IllegalArgumentException("Factorial input has to be between 1 and 10");
    }

    // rest of your code....
}