Java : Sort integer array without using Arrays.sort()

Question

This is the instruction in one of the exercises in our Java class. Before anything else, I would like to say that I 'do my homework' and I'm not just being lazy asking someone on Stack Overflow to answer this for me. This specific item has been my problem out of all the other exercises because I've been struggling to find the 'perfect algorithm' for this.

Write JAVA program that will input 10 integer values and display either in ascending or descending order. Note: Arrays.sort() is not allowed.

This is the code I have come up with, it works but it has one obvious flaw. If I enter the same value twice or more, for example:

5, 5, 5, 4, 6, 7, 3, 2, 8, 10

Only one of the three 5s entered would be counted and included in the output. The output I get (for the ascending order) is:

2 3 4 5 0 0 6 7 8 10.

import java.util.Scanner;

public class Exer3AscDesc
{
    public static void main(String args[])
    {
        Scanner scan = new Scanner(System.in);
        int tenNums[]=new int[10], orderedNums[]=new int[10];
        int greater;
        String choice;

        //get input
        System.out.println("Enter 10 integers : ");
        for (int i=0;i<tenNums.length;i++)
        {
            System.out.print(i+1+"=> ");
            tenNums[i] = scan.nextInt();
        }
        System.out.println();

        //imperfect number ordering algorithm
        for(int indexL=0;indexL<tenNums.length;indexL++)
        {
            greater=0;
            for(int indexR=0;indexR<tenNums.length;indexR++)
            {
                if(tenNums[indexL]>tenNums[indexR])
                {
                    greater++;
                }
            }
            orderedNums[greater]=tenNums[indexL];
        }

        //ask if ascending or descending
        System.out.print("Display order :\nA - Ascending\nD - Descending\nEnter your choice : ");
        choice = scan.next();

        //output the numbers based on choice
        if(choice.equalsIgnoreCase("a"))
        {
            for(greater=0;greater<orderedNums.length;greater++)
            {
                System.out.print(orderedNums[greater]+" ");
            }
        }
        else if(choice.equalsIgnoreCase("d"))
        {
            for(greater=9;greater>-1;greater--)
            {
                System.out.print(orderedNums[greater]+" ");
            }
        }
    }
}

Would it be cheating to create a PriorityQueue, add all the items, and then repeatedly poll until the queue is empty? — Jim Mischel
– Jim Mischel, Commented Aug 31, 2016 at 11:39

gvlasov · Accepted Answer · 2015-02-28 19:25:13Z

34

Simple sorting algorithm Bubble sort:

public static void main(String[] args) {
    int[] arr = new int[] { 6, 8, 7, 4, 312, 78, 54, 9, 12, 100, 89, 74 };

    for (int i = 0; i < arr.length; i++) {
        for (int j = i + 1; j < arr.length; j++) {
            int tmp = 0;
            if (arr[i] > arr[j]) {
                tmp = arr[i];
                arr[i] = arr[j];
                arr[j] = tmp;
            }
        }
    }
}

edited Feb 28, 2015 at 19:25

gvlasov

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answered Feb 28, 2015 at 19:02

Yoga

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gvlasov · Accepted Answer · 2015-02-28 19:29:09Z

6

You can find so many different sorting algorithms in internet, but if you want to fix your own solution you can do following changes in your code:

Instead of:

 orderedNums[greater]=tenNums[indexL];

you need to do this:

while (orderedNums[greater] == tenNums[indexL]) {
     greater++;
}
orderedNums[greater] = tenNums[indexL];

This code basically checks if that particular index is occupied by a similar number, then it will try to find next free index.

Note: Since the default value in your sorted array elements is 0, you need to make sure 0 is not in your list. otherwise you need to initiate your sorted array with an especial number that you sure is not in your list e.g: Integer.MAX_VALUE

edited Feb 28, 2015 at 19:29

gvlasov

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answered Nov 25, 2012 at 6:13

mhshams

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Thanks a lot mohammad, I spent too much time trying to formulate different algorithms, I didn't realize how simple it was that I just needed to add that one line to my code.

ransan32
– ransan32

11/25/2012 15:53:37
Commented Nov 25, 2012 at 15:53

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Premkumar Manipillai · Accepted Answer · 2018-02-22 09:40:16Z

4

Simple way :

int a[]={6,2,5,1};
            System.out.println(Arrays.toString(a));
             int temp;
             for(int i=0;i<a.length-1;i++){
                 for(int j=0;j<a.length-1;j++){
                     if(a[j] > a[j+1]){   // use < for Descending order
                         temp = a[j+1];
                         a[j+1] = a[j];
                         a[j]=temp;
                     }
                 }
             }
            System.out.println(Arrays.toString(a));

    Output:
    [6, 2, 5, 1]
    [1, 2, 5, 6]

answered Feb 22, 2018 at 9:40

Premkumar Manipillai

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in the inner loop you have not used variable i at all

Vadiraj Purohit
– Vadiraj Purohit

03/01/2020 04:24:50
Commented Mar 1, 2020 at 4:24

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Jesse · Accepted Answer · 2013-04-17 00:55:25Z

Here is one simple solution

public static void main(String[] args) {        
        //Without using Arrays.sort function
        int i;
        int nos[] = {12,9,-4,-1,3,10,34,12,11};
        System.out.print("Values before sorting: \n");
        for(i = 0; i < nos.length; i++)
            System.out.println( nos[i]+"  ");               
        sort(nos, nos.length);
        System.out.print("Values after sorting: \n");       
        for(i = 0; i <nos.length; i++){
            System.out.println(nos[i]+"  ");
        }
    }

    private static void sort(int nos[], int n) {
     for (int i = 1; i < n; i++){
          int j = i;
          int B = nos[i];
          while ((j > 0) && (nos[j-1] > B)){
            nos[j] = nos[j-1];
            j--;
          }
          nos[j] = B;
        }
    }

And the output is:

Values before sorting:

12
9
-4
-1
3
10
34
12
11

Values after sorting:

-4
-1
3
9
10
11
12
12
34

gvlasov · Accepted Answer · 2015-02-28 19:30:10Z

1

I would recommend looking at Selection sort or Insertion sort if you aren't too worried about performance. Maybe that will give you some ideas.

edited Feb 28, 2015 at 19:30

gvlasov

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answered Nov 25, 2012 at 5:36

awolfe91

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Narendra · Accepted Answer · 2018-01-12 18:37:48Z

int x[] = { 10, 30, 15, 69, 52, 89, 5 };
    int max, temp = 0, index = 0;
    for (int i = 0; i < x.length; i++) {
        int counter = 0;
        max = x[i];
        for (int j = i + 1; j < x.length; j++) {

            if (x[j] > max) {
                max = x[j];
                index = j;
                counter++;
            }

        }
        if (counter > 0) {
            temp = x[index];
            x[index] = x[i];
            x[i] = temp;
        }
    }
    for (int i = 0; i < x.length; i++) {
        System.out.println(x[i]);
    }

realPK · Accepted Answer · 2016-07-28 21:30:34Z

0

Bubble sort can be used here:

 //Time complexity: O(n^2)
public static int[] bubbleSort(final int[] arr) {

    if (arr == null || arr.length <= 1) {
        return arr;
    }

    for (int i = 0; i < arr.length; i++) {
        for (int j = 1; j < arr.length - i; j++) {
            if (arr[j - 1] > arr[j]) {
                arr[j] = arr[j] + arr[j - 1];
                arr[j - 1] = arr[j] - arr[j - 1];
                arr[j] = arr[j] - arr[j - 1];
            }
        }
    }

    return arr;
}

answered Jul 28, 2016 at 21:30

realPK

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mehrdad khosravi · Accepted Answer · 2016-08-31 11:38:11Z

Array Sorting without using built in functions in java ......just make new File unsing this name -> (ArraySorting.java) ..... Run the Project and Enjoy it !!!!!

    import java.io.*;
    import java.util.Arrays;
    import java.util.Scanner;
    public class ArraySorting 
    {
    public static void main(String args[])
    {
      int temp=0;   
      Scanner user_input=new Scanner(System.in);
       System.out.println("enter Size of Array...");
    int Size=user_input.nextInt();

    int[] a=new int[Size];
    System.out.println("Enter element Of an Array...");
    for(int j=0;j<Size;j++)
    {
        a[j]=user_input.nextInt();
    }     
    for(int index=0;index<a.length;index++)
    {
        for(int j=index+1;j<a.length;j++)
        {
             if(a[index] > a[j] ) 
             {
                 temp = a[index];
                 a[index] = a[j];
                 a[j] = temp;
             }
        }
    }
    System.out.print("Output is:- ");
    for(int i=0;i<a.length;i++)
    {
        System.out.println(a[i]);
    }

}
}

Suraj Rao · Accepted Answer · 2018-11-20 08:51:59Z

int[] arr = {111, 111, 110, 101, 101, 102, 115, 112}; /* for ascending order */ System.out.println(Arrays.toString(getSortedArray(arr))); /*for descending order */ System.out.println(Arrays.toString(getSortedArray(arr))); private int[] getSortedArray(int[] k){ int localIndex =0; for(int l=1;l<k.length;l++){ if(l>1){ localIndex = l; while(true){ k = swapelement(k,l); if(l-- == 1) break; } l = localIndex; }else k = swapelement(k,l); } return k; } private int[] swapelement(int[] ar,int in){ int temp =0; if(ar[in]<ar[in-1]){ temp = ar[in]; ar[in]=ar[in-1]; ar[in-1] = temp; } return ar; } private int[] getDescOrder(int[] byt){ int s =-1; for(int i = byt.length-1;i>=0;--i){ int k = i-1; while(k >= 0){ if(byt[i]>byt[k]){ s = byt[k]; byt[k] = byt[i]; byt[i] = s; } k--; } } return byt; }

output:-
ascending order:- 101, 101, 102, 110, 111, 111, 112, 115

descending order:- 115, 112, 111, 111, 110, 102, 101, 101

Uri Agassi · Accepted Answer · 2015-03-10 15:32:37Z

class Sort 
{
    public static void main(String[] args) 
    {
    System.out.println("Enter the range");
    java.util.Scanner sc=new java.util.Scanner(System.in);
    int n=sc.nextInt();
    int arr[]=new int[n];
    System.out.println("Enter the array values");
    for(int i=0;i<=n-1;i++)
    {
        arr[i]=sc.nextInt();

    }
    System.out.println("Before sorting array values are");
     for(int i=0;i<=n-1;i++)
    {
     System.out.println(arr[i]);
       }
    System.out.println();
    for(int pass=1;pass<=n;pass++)
    {
    for(int i=0;i<=n-1;i++)
    {
        if(i==n-1)
        {
            break;
        }
        int temp;
        if(arr[i]>arr[i+1])
        {

            temp=arr[i];
            arr[i]=arr[i+1];
            arr[i+1]=temp;
        }
    }
    }

    System.out.println("After sorting array values are");
       for(int i=0;i<=n-1;i++)
    {
     System.out.println(arr[i]);
       }

}
}

This is my logic go through it

saroj kumar patnaik
– saroj kumar patnaik

03/10/2015 15:18:00
Commented Mar 10, 2015 at 15:18 — saroj kumar patnaik
– saroj kumar patnaik, Commented Mar 10, 2015 at 15:18

Srinivas Nellore · Accepted Answer · 2016-06-06 06:36:39Z

here is the Sorting Simple Example try it

public class SortingSimpleExample {

    public static void main(String[] args) {
        int[] a={10,20,1,5,4,20,6,4,2,5,4,6,8,-5,-1};
              a=sort(a);
        for(int i:a)
              System.out.println(i);
}
    public static int[] sort(int[] a){

        for(int i=0;i<a.length;i++){
            for(int j=0;j<a.length;j++){
                int temp=0;
                if(a[i]<a[j]){
                    temp=a[j];                  
                    a[j]=a[i];                  
                    a[i]=temp;      
                  }         
                }
        }
        return a;

    }
}

Khalil M · Accepted Answer · 2016-12-15 21:21:15Z

-2

This will surely help you.

int n[] = {4,6,9,1,7};

    for(int i=n.length;i>=0;i--){
        for(int j=0;j<n.length-1;j++){
            if(n[j] > n[j+1]){
                swapNumbers(j,j+1,n);
            }
        }

    }
    printNumbers(n);
}
private static void swapNumbers(int i, int j, int[] array) {

    int temp;
    temp = array[i];
    array[i] = array[j];
    array[j] = temp;
}

private static void printNumbers(int[] input) {

    for (int i = 0; i < input.length; i++) {
        System.out.print(input[i] + ", ");
    }
    System.out.println("\n");
}

edited Dec 15, 2016 at 21:21

Khalil M

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answered Oct 9, 2015 at 14:31

Krishna Kumar Chourasiya

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