How to print unique elements in an array by returning the array?

Question

I have this following code snippet and want to print unique elements of the array a[]
I want to print the result in main method by returning an array. Following code returning a result- 10 20 21 34 0 56 65 0 76 67 45 55 0 99 23 0 0 0 12
Can anyone please tell me where I am doing the mistake?

public class uniqueElements {
static int[] newArr=null;
public static void main(String[] args){
 int[] a=new int[] {10,20,21,34,21,56,65,65,76,67,45,55,10,99,23,67,99,34,12};
 int[] b=uniqueElements(a);
 for(int i:b){
     System.out.print(i+" ");
 }

 }

private static int[] uniqueElements(int[] arr) {
newArr=new int[arr.length];
for(int i=0;i<arr.length;i++){
    boolean isDistinct = false;
    for(int j=0;j<i;j++){
        if(arr[i] == arr[j]){
            isDistinct = true;
            break;
        }
    }
    if(!isDistinct){

        newArr[i]=arr[i];

    }
}
return newArr;
}
}

What do you expect to happen? That result seems to be exactly what your code would do. (Nevermind the various other problems with your code style.) — Alexis King
– Alexis King, Commented Jan 23, 2015 at 20:21
As I said, I want to print unique elements of the array. I want to know where I am doing the mistake in this code. I am storing the unique elements in the array newArr and returning it. — Piku
– Piku, Commented Jan 23, 2015 at 20:25

martintama · Accepted Answer · 2015-01-23 20:27:22Z

2

You can also use a Set if you only want unique values:

http://docs.oracle.com/javase/7/docs/api/java/util/Set.html

Set<Integer> b = new HashSet<Integer>(Arrays.asList(a));

But if you still want to do it manually, you can do as the other people said.

answered Jan 23, 2015 at 20:27

martintama

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Comments

Eran · Accepted Answer · 2015-01-23 20:29:49Z

You copy distinct values from the input array to the same index in the output array, which means the indices of duplicate values will remain 0, which is what you see in your output.

A possible solution :

private static int[] uniqueElements(int[] arr) 
{
    newArr=new int[arr.length];
    int count = 0;
    for(int i=0;i<arr.length;i++){
        boolean isDistinct = false;
        for(int j=0;j<i;j++){
            if(arr[i] == arr[j]){
                isDistinct = true;
                break;
            }
        }
        if(!isDistinct) {
            newArr[count]=arr[i];
            count++;
        }
    }
    return newArr;
}

This would return an array whose first count elements are the unique values of the input array. It would still have some 0s at the end. To eliminate these 0s, you can create a new array of count elements and copy the first count elements of newArr to it.

BTW, your isDistinct flag has the exact opposite meaning to what its name implies. You set it to true when you find a duplicate.

Toby Derrum · Accepted Answer · 2015-01-23 20:50:48Z

1

The easiest way to remove repeated elements is to add the contents to a Set (which will not allow duplicates) and then add the Set back to the ArrayList

public class uniqueElements {
static int[] newArr=null;
public static void main(String[] args){
 int[] a=new int[] {10,20,21,34,21,56,65,65,76,67,45,55,10,99,23,67,99,34,12};
 int[] b=uniqueElements(a);
 for(int i:b){
     System.out.print(i+" ");
 }

 }

private static int[] uniqueElements(int[] a) {

// add elements to a, including duplicates
HashSet hs = new HashSet();
hs.addAll(a);
a.clear();
a.addAll(hs);
return a;
}
}

edited Jan 23, 2015 at 20:50

answered Jan 23, 2015 at 20:34

Toby Derrum

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