How to input different numbers with Java

Question

The question is to Write a program that sorts three integers. The integers are entered from the input dialogs and stored in variables num1, num2, and num3, respectively. The program sorts the numbers so that num1 <= num2 <= num3.

actually I do that but the result is available only to 1 ,2 and 3 numbers !

When I enter any different number it doesn't show me the result I want it !

here is my code..

import javax.swing.JOptionPane;

public class number order {

 public static void main(String[] args) {

 int num1;
 int num2;
 int num3;

    String n = JOptionPane.showInputDialog(null, "input NUM 1 " );
           num1 = Integer.parseInt(n);
    String u = JOptionPane.showInputDialog(null, "input NUM 2 " );
         num2 = Integer.parseInt(u);
    String m = JOptionPane.showInputDialog(null, "input NUM 3 " );
        num3 = Integer.parseInt(m);

    if (num1<=num2&& num2<=num3)
        System.out.println( num1+"<="+ num2+"<="+num3 );
    if(num2<=num1&&num1<=num3)
        System.out.println(num2+"<="+num1+"<="+num3);
    if (num3<=num1&&num1<=num2)
        System.out.println(num3+"<="+num1+"<="+num2);


    // TODO code application logic here
 }
}

Could you rephrase your question? Do you mean that the program can only accept 3 numbers as input (and not, e.g., four or five)? Or do you mean that the "sorting" works only for certain inputs but not for others? — tobias_k
– tobias_k, Commented Oct 18, 2013 at 8:06
I mean the sorting works only for certain inputs which are ( 1,2 and 3 ) .. — Eng .
– Eng ., Commented Oct 18, 2013 at 8:13
Just to make this clear, if you input, say, 6, 2, and 9, the sorting does not work? — tobias_k
– tobias_k, Commented Oct 18, 2013 at 8:14
yes exactly and when I input 1,2and 3 it works .. 1<=2<=3 .. but other numbers are not — Eng .
– Eng ., Commented Oct 18, 2013 at 8:17
Could you check this again? There's nothing in your code which would restrict input to integers 1,2,3. In fact, 6,2,9 (in this order) should be sorted just fine. — tobias_k
– tobias_k, Commented Oct 18, 2013 at 8:47

tobias_k · Accepted Answer · 2013-10-18 09:25:07Z

The problem is that you check only three out of six possible arrangements of those three numbers. Also note that, even for those three, you are not actually sorting the numbers, but only printing them in sorted order, i.e., you are never reassigning the variables num1, num2, and num3.

Just as an alternative to checking all the possible arrangements of the three numbers or implementing a full sorting algorithm, you can also compare and swap pairs of numbers. This way, you get away with far fewer comparisons while still being able to sort all permutations of three numbers.

if num1 > num2, swap num1 and num2
if num2 > num3, swap num2 and num3
if num1 > num2, swap num1 and num2 again

After those three swaps, the numbers are in sorted order.

Of course, if you have more than three numbers this gets impractical, and you should rather implement a full sorting algorithm (for exercise) or go with one of the builtins, like Arrays.sort (for real life).

Joachim Weiß · Accepted Answer · 2013-10-18 08:14:09Z

0

You haven't checked all cases:

1 < 2 < 3
1 < 3 < 2
2 < 1 < 3
2 < 3 < 1
3 < 1 < 2
3 < 2 < 1

However to check all case like that is not to useful. an sorted array would be more easy:

int arr[3]={num1,num2,num3}
java.utils.Arrays.sort(arr);
println(arr[0] + "<=" + arr[1] + "<=" + arr[2]);

answered Oct 18, 2013 at 8:14

Joachim Weiß

4072 silver badges12 bronze badges

Comments

Pankaj Sharma · Accepted Answer · 2013-10-18 08:22:20Z

0

int[] all = new int[]();
num1 = Integer.parseInt(n);
all[0] = num1;
num2 = Integer.parseInt(u);
all[1] = num1;
num3 = Integer.parseInt(m);
all[2] = num1;
for(int i=0; i <all.length ;i++){
  if(i!=0 && all[i]< all[i-1]){
     temp = all[i-1];
     all[i-1] = all[i];
     all[i] = temp; 
  }
}

all will have sorted array
or more simply

2 steps
1)add all to a array.
2) call Arrays.sort(array)

answered Oct 18, 2013 at 8:22

Pankaj Sharma

1,8531 gold badge18 silver badges22 bronze badges

1 Comment

Eng . Over a year ago

Is my code write ? I just need to work in JOptionPane method ! and thanks so much ..

k_zaur_k · Accepted Answer · 2013-10-18 08:58:10Z

0

how about

int min = min(min(num1,num2),num3);
int max = max(max(num1,num2),num3);
int mid = num1 + num2 + num3 - min - max;
System.out.println(min+"<="+mid+"<="+max);

answered Oct 18, 2013 at 8:58

k_zaur_k

4002 gold badges6 silver badges16 bronze badges

Comments

Ashish · Accepted Answer · 2013-10-18 09:08:38Z

0

if(num1 < num2){ if(num3 < num1) System.out.println("num2 > num1 > num3");}
else{ if(num2 < num3) System.out.println("num3 > num2 > num1"); else System.out.println("num1 > num2 > num3");}

Sort in descending order.

Hope it helps.

answered Oct 18, 2013 at 9:08

Ashish

7451 gold badge6 silver badges15 bronze badges

Collectives™ on Stack Overflow

How to input different numbers with Java

5 Answers 5

Comments

Comments

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related