CODEWITHSUNDEEP
javaarrayssortingintegerint
ransan32
ransan32

Reputation: 235

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

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]+" ");
            }
        }
    }
}

Upvotes: 12

Views: 188846

Answers (12)

Shubham Gaur
Shubham Gaur

Reputation: 1

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

Upvotes: 0

Premkumar Manipillai
Premkumar Manipillai

Reputation: 2151

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]

Upvotes: 4

Narendra
Narendra

Reputation: 137

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]);
    }

Upvotes: 1

Krishna Kumar Chourasiya
Krishna Kumar Chourasiya

Reputation: 2136

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");
}

Upvotes: -2

Adeel Asghar
Adeel Asghar

Reputation: 1

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]);
    }

}
}

Upvotes: 0

realPK
realPK

Reputation: 2961

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;
}

Upvotes: 0

Srinivas Nellore
Srinivas Nellore

Reputation: 9

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;

    }
}

Upvotes: -1

saroj kumar patnaik
saroj kumar patnaik

Reputation: 1

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]);
       }

}
}

Upvotes: -1

awolfe91
awolfe91

Reputation: 1647

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.

Upvotes: 1

mhshams
mhshams

Reputation: 16952

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

Upvotes: 6

Yoga
Yoga

Reputation: 551

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;
            }
        }
    }
}

Upvotes: 34

Gopinath
Gopinath

Reputation: 21

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

Upvotes: 2

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