How to remove duplicates from a list using an auxi

Question

I am trying to remove duplicates from a list by creating a temporary array that stores the indices of where the duplicates are, and then copies off the original array into another temporary array while comparing the indices to the indices I have stored in my first temporary array.

public void removeDuplicates()
{
    double tempa [] = new double [items.length];
    int counter = 0;
    for ( int i = 0; i< numItems ; i++)
    {
        for(int j = i + 1; j < numItems; j++)
        {
            if(items[i] ==items[j])
            {
                tempa[counter] = j;
                counter++;

            }
        }
    }

    double tempb [] = new double [ items.length];
    int counter2 = 0;
    int j =0;
    for(int i = 0; i < numItems; i++)
    {
        if(i != tempa[j])
        {
            tempb[counter2] = items[i];
            counter2++;

        }
        else
        {
            j++;

        }
    }

    items = tempb;
    numItems = counter2;
}

and while the logic seems right, my compiler is giving me an arrayindexoutofbounds error at

tempa[counter] = j;

I don't understand how counter could grow to above the value of items.length, where is the logic flaw?

Answer 1

Instead of doing it in array, you can simply use java.util.Set.

Here an example:

public static void main(String[] args)
{
    Double[] values = new Double[]{ 1.0, 2.0, 2.0, 2.0, 3.0, 10.0, 10.0 };
    Set<Double> singleValues = new HashSet<Double>();

    for (Double value : values)
    {
        singleValues.add(value);
    }
    System.out.println("singleValues: "+singleValues);
    // now convert it into double array
    Double[] dValues = singleValues.toArray(new Double[]{});
}

Answer 2

import java.util.HashSet;

import sun.security.util.Length;


public class arrayduplication {
public static void main(String[] args) {
        int arr[]={1,5,1,2,5,2,10};
        TreeSet< Integer>set=new TreeSet<Integer>();
        for(int i=0;i<arr.length;i++){
            set.add(Integer.valueOf(arr[i]));
        }
        System.out.println(set);


    }

}

Answer 3

You have already used num_items to bound your loop. Use that variable to set your array size for tempa also.

double tempa [] = new double [num_items];

Answer 4

You are making things quite difficult for yourself. Let Java do the heavy lifting for you. For example LinkedHashSet gives you uniqueness and retains insertion order. It will also be more efficient than comparing every value with every other value.

double [] input = {1,2,3,3,4,4};
Set<Double> tmp = new LinkedHashSet<Double>();
for (Double each : input) {
    tmp.add(each);
}
double [] output = new double[tmp.size()];
int i = 0;
for (Double each : tmp) {
    output[i++] = each;
}
System.out.println(Arrays.toString(output));

Answer 5

You can use a set for removing multiples.

Answer 6

Here's another alternative without the use of sets, only primitive types:

public static double [] removeDuplicates(double arr[]) {
    double [] tempa = new double[arr.length];
    int uniqueCount = 0;
    for (int i=0;i<arr.length;i++) {
        boolean unique = true;
        for (int j=0;j<uniqueCount && unique;j++) {
            if (arr[i] == tempa[j]) {
                unique = false;
            }
        }
        if (unique) {
            tempa[uniqueCount++] = arr[i];
        }
    }

    return Arrays.copyOf(tempa,  uniqueCount);
}

It does require a temporary array of double objects on the way towards getting your actual result.