如何找到Java中输入的最小序列数

Question

I am devoloping an application to find the minimum of all the numbers entered .It accepts the numbers from the dialog box and when the user enters 0 it displays the minimum of all the numbers.But i dont need the 0 but the minimum of the numbers that preceeded it.

My code is as Follows:

    try {
        int a, c = 0, d = 0;

        do {
            a = Integer.parseInt(jOptionPane1.showInputDialog(null, "please enter the number"));

            c = Math.min(a, d);
            if (a != 0) //since a=0 will be excecuted one time
            {
                d = c;
            }
        } while (a != 0);

        lb2.setText("Minimum of the numbers is " + d);



    } catch (Exception e) {
        jOptionPane1.showMessageDialog(this, "Some thing went wrong");
    }

I know that it gives me 0 because the minimim of the numbers entered is zero and if i enter a number less than 0 (ie a negative number)it gives me the correct answer .I think the problem is also due to the initialisation that c=0.

Now i need a method to find the minimum without using any arrays and it should be simple and easy.(most helpful if you use Math.min itself)

Any help Appreciated.

Answer 1

只需将您的初始化更改为将d设置为Integer.MAX_VALUE 。

Answer 2

I have an advice for your code, that you should make every variable names make sense. Maybe your code is small, but it will affect your habit, and when you work in large project, your habit will affect you so much :)

You should change initialize part of d = 0 to d = Integer.MAX_VALUE . Here is a new code :

try {
        int inputNumber = 0, min= Integer.MAX_VALUE;
        // int c : you don't need this variable

        do {
            inputNumber = Integer.parseInt(jOptionPane1.showInputDialog(null, "please enter the number"));

            if (inputNumber != 0) //since a=0 will be excecuted one time
            {
                min= inputNumber;
            }
        } while (inputNumber != 0);

        lb2.setText("Minimum of the numbers is " + min);



    } catch (Exception e) {
        jOptionPane1.showMessageDialog(this, "Some thing went wrong");
    }

This new code is make more sense ?

And, Why you Must change initialize to min= Integer.MAX_VALUE ? For example, you initialize like this : min = 10; . And at first time when someone type 15, you program will see : Oh, 15>10, so this is not the min value. But in fact, 15 is the first value of input and it should be the min value. Your program will be wrong until someone type a number less than 10.

Compare to your code, because you initialize d=0 . when you type d=1 , ops, 1>0, this is not min value (like above example). And everything will true only when you type some numbers < 0.

Al the problem above happen, because although user types any number, the min is the initialize number. (the number that user doesn't type). And that why you set your min value to something REALLY REALLY big. So, the FIRST TIME you type some numbers, It's ALREADY the SMALLEST .

Because machine not like us, doesn't have "infinity", so we must set it the biggest value possible. And because d is an int, so the biggest value of int is Integer.MAX_VALUE . In fact, you can type number for d. I don't remember exactly, but the biggest value for integer in range 32000 (2^31).