Sorting of integer using ArrayList <Integer>

Question

Sorting method should returns a string with a comma separated list of integers from the array, sorted in ascending order. My code does not sort for the size of 4 or more. I am also not getting anything in console for return type without system.out.println(); Any suggestion /help is appreciated.



import java.util.ArrayList;

public class Sorting {  
    public static void main(String[] args) 
    {
        ArrayList<Integer> alInput = makeArray(5);
        ArraySorter(alInput);

    }
    public static String ArraySorter (ArrayList<Integer> alSort) 
    {
        if (alSort.size()<=1)
        {
            return String.valueOf(alSort.get(0)) ;
        }
        int iMin = alSort.get(0);
        int iMinIndex = 0;
        for (int i = 0;i<alSort.size();i++ )
        {
            if (alSort.get(i)<iMin)
            {
                iMin = alSort.get(i);
                iMinIndex = i;
            }
        }
        alSort.remove(iMinIndex);
        System.out.println(String.valueOf(iMin) + ArraySorter (alSort));
        return String.valueOf(iMin) + ArraySorter (alSort);
    }

    public static ArrayList<Integer> makeArray(int iSize)
    {
        ArrayList<Integer> alNum = new ArrayList<Integer>();

        for (int i = 0; i<iSize;i++)
        {
            alNum.add((int )(Math.random() * 100 + 1));
        }
        System.out.println(alNum);
        return alNum;
    }
}

Use simple for loop to get through all elements in array, 1 boolean variable that will state if loop is sorted or not and 1 temp variable to save element from an array while you do swapping if element[i] is greater than element[i + 1] (if you want it ascending). — user7451361
– user7451361, Commented Nov 8, 2017 at 22:30
Have you searched this stackoverflow.com/search?q=Sorting+ArrayList+java stackoverflow.com/search?q=quick+sort+algorithms+java — B-GangsteR
– B-GangsteR, Commented Nov 8, 2017 at 23:54

Rishabh Dugar · Accepted Answer · 2017-11-09 04:13:06Z

1

Why dont you try this easy way,

alInput.sort(null);
        String myString = "";
        for (int i=0;i<alInput.size();i++){
            myString  = myString + String.valueOf(alInput.get(i)) + ",";
        }
        myString= myString.substring(0,myString.length()-1);
        System.out.println(myString);

answered Nov 9, 2017 at 4:13

Rishabh Dugar

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2 Comments

Apurva Gandhi Over a year ago

Thanks Rishabh! This works but I was not allowed to use .sort for this project. We needed to use recursion in the code. Thanks any way buddy! @Rishabh Dugar

Rishabh Dugar Over a year ago

Oh , that was not mentioned in problem which i why i answered . :)

Dragoş Bercea · Accepted Answer · 2017-11-09 04:05:56Z

0

Give this a try: I have added comments explaining what I have changed for the fixe(s). Hope it helps :)

import java.util.ArrayList;

public class Sorting {  
    public static void main(String[] args) 
    {
        ArrayList<Integer> alInput = makeArray(5);

        // now you can get the sorted string :)
        String sortedResult = ArraySorter(alInput);
        System.out.println("The sorted array is : " + sortedResult);

    }
    public static String ArraySorter (ArrayList<Integer> alSort) 
    {
        int iMin = alSort.get(0);
        int iMinIndex = 0;
        for (int i = 0;i<alSort.size();i++ )
        {
            if (alSort.get(i)<iMin)
            {
                iMin = alSort.get(i);
                iMinIndex = i;
            }
        }
        alSort.remove(iMinIndex);

        // added this one in myself, this is your base case for preventing a stack overflow:
        if (alSort.size() == 0) {
            return String.valueOf(iMin);
        }

        return String.valueOf(iMin) + "," + ArraySorter (alSort); // changed from : String.valueOf(iMin) + ArraySorter (alSort);
    }

    public static ArrayList<Integer> makeArray(int iSize)
    {
        ArrayList<Integer> alNum = new ArrayList<Integer>();

        for (int i = 0; i<iSize;i++)
        {
            alNum.add((int )(Math.random() * 100 + 1));
        }
        System.out.println(alNum);
        return alNum;
    }
}

answered Nov 9, 2017 at 4:05

Dragoş Bercea

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2 Comments

Apurva Gandhi Over a year ago

Thanks buddy! this is what I was missing! Thanks for your help man! @Dragoş Bercea

Dragoş Bercea Over a year ago

No worries mate, It was pretty fun to debug.

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