ascending sort without using arrays in java

Question

This code has to enter 5 integers and then display them in ascending and descending orders without using arrays. But it gives me only the biggest number. How can i do it without arrays??

package assignment;

import javax.swing.JOptionPane;

public class q1 {

    public static void main(String[] args) {

        String a = JOptionPane.showInputDialog(null, "Please Enter a number", "Input", JOptionPane.QUESTION_MESSAGE);
        int number1 = Integer.parseInt(a);
        String b = JOptionPane.showInputDialog(null, "Please Enter a number", "Input", JOptionPane.QUESTION_MESSAGE);
        int number2 = Integer.parseInt(b);
        String c = JOptionPane.showInputDialog(null, "Please Enter a number", "Input", JOptionPane.QUESTION_MESSAGE);
        int number3 = Integer.parseInt(c);
        String d = JOptionPane.showInputDialog(null, "Please Enter a number", "Input", JOptionPane.QUESTION_MESSAGE);
        int number4 = Integer.parseInt(d);
        String e = JOptionPane.showInputDialog(null, "Please Enter a number", "Input", JOptionPane.QUESTION_MESSAGE);
        int number5 = Integer.parseInt(e);

        if (number1 > number2 && number1 > number3 && number1 > number4 && number1 > number5) {
            System.out.print(number1 + ",");
        } else if (number2 > number1 && number2 > number3 && number2 > number4 && number2 > number5) {
            System.out.print(number2 + ",");
        } else if (number3 > number1 && number3 > number2 && number3 > number4 && number3 > number5) {
            System.out.print(number3 + ",");
        } else if (number4 > number1 && number4 > number2 && number4 > number3 && number4 > number5) {
            System.out.print(number4 + ",");
        } else if (number5 > number1 && number5 > number2 && number5 > number3 && number5 > number4) {
            System.out.print(number5 + ",");
        }

    }
}

Does it say "without using arrays", or does it say "without using `Arrays`"? — khelwood, Sep 29 '14 at 16:42
Well, you're using else if logic, only one if statement would get executed ever. — proulxs, Sep 29 '14 at 16:49
You want a sorting network. http://jgamble.ripco.net/cgi-bin/nw.cgi?inputs=5&algorithm=best&output=macro — BonzaiThePenguin, Sep 29 '14 at 16:49
no Arrays, how about a priority queue? or, keep iterating over your input boxes and each time you find the remaining maximum, set it to some impossible value that tells you it's been grabbed. — user1269942, Sep 29 '14 at 17:02
btw, figuring out your own homework will help more in the long run than coming here! — user1269942, Sep 29 '14 at 17:03
Take it easy guys. Though there are lots of tutorials and explanations on sorting, there aren't really any for doing sorts outside of methods and using data stuctures, which he isn't allowed to use at the moment. — MarGar, Sep 29 '14 at 17:12

MarGar · Accepted Answer · 2014-09-29T17:17:23.200

3

Oddly I couldn't really find much on sorting outside of arrays and lists so I will assist you.

Code:

  int temp;

for (int count = 0; count < 4; count++)
{
    if ( number2 < number1)
    {
       temp = number1;
       number1 = number2;
       number2 = temp;
    }

    if ( number3 < number2)
    {
       temp = number2;
       number2 = number3;
       number3 = temp;
    }

    if ( number4 < number3)
    {
       temp = number3;
       number3 = number4;
       number4 = temp;
    }

    if ( number5 < number4)
    {
       temp = number4;
       number4 = number5;
       number5 = temp;
    }
}

System.out.println( number1 + " " + number2 + " " + number3 + " " + number4 + " " + number5);
System.out.println( number5 + " " + number4 + " " + number3 + " " + number2 + " " + number1);

Sorting will get easier as you learn some of the algorithms and how they work. Until then, I hope this helps.

The basic idea is to check each pair and swap the values. You will need to do this multiple times in order to get the last value to the beginning of your list of values.

edited Sep 29 '14 at 17:17

answered Sep 29 '14 at 16:59

MarGar

465
5
12

2

This seems more appropriate, doing it 118 times like the other answer suggests is nuts :P – RockOnRockOut Sep 29 '14 at 17:06
Thanks for your helpful comment Mar, your code works. – beshr Sep 29 '14 at 17:19
My answer was more of a joke. Though it would work if you took the time to write out all 120 permutations. This is obviously a more elegant answer. – dberm22 Sep 29 '14 at 19:55
Hey, if it works then it IS an answer! I thought it was funny :). – MarGar Sep 29 '14 at 20:01

score 2 · Answer 2 · answered Sep 29 '14 at 16:59

Easy. It can be done in 120 lines.

if(number1>=number2 && number2>=number3 && number3>=number4 && number4>=number5)
    System.out.print(number1 + "," + number2 + "," + number3 + "," + number4 + "," + number5);
else if (number1>=number2 && number2>=number3 && number3>=number5 && number5>=number4)
    System.out.print(number1 + "," + number2 + "," + number3 + "," + number5 + "," + number4);
else if ...

Just do this 118 more times for every permutation of 5P5 :P

Brute force to the max.

Enjoy doing this with 10 integers.

Sunil Chawla · Answer 3 · 2014-09-29T18:18:44.603

I don't think the logic you are using is correct.

I suppose the code for arranging 5 int variables in decending order will be

Assume that the variables are initialized and assigned

int max;
int temp;
for(int i=1 ; i<=4 ; i++)
{
    if(i==1)
    {
        max=n1;
    }
    if(i==2)
    {
        max=n2;
    }
    if(i==3)
    {
        max=n3;
    }
    if(i==4)
    {
        max=n4;
    }

    for(int j=i+1 ; j<=5 ; j++)
    {
        if(j<=2)
        {
            if(n2>max)
            {
                temp=n2;
                n2=max;
                max=temp;
            }
        }

        if(j<=3)
        {
            if(n3>max)
            {
                temp=n3;
                n3=max;
                max=temp;
            }
        }

        if(j<=4)
        {
            if(n4>max)
            {
                temp=n4;
                n4=max;
                max=temp;
            }
        }

        if(j<=5)
        {
            if(n5>max)
            {
                temp=n5;
                n5=max;
                max=temp;
            }
        }   

    }
    if(i==1)
    {
        n1=max;
    }
    if(i==2)
    {
        n2=max;
    }
    if(i==3)
    {
        n3=max;
    }
    if(i==4)
    {
        n4=max;
    }
}
System.out.print(n1 + "," + n2 + "," + n3 + "," + n4 + "," + n5);

Well..I am basically using Selection Sort here.

Be careful, it's easy to get a down vote, especially when your code doesn't work. +1 for an alternate solution. — MarGar, Sep 29 '14 at 18:23

ascending sort without using arrays in java

3 Answers3

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