Pulling distinct values from a array in java

In Java 8

Stream< T > distinct()

Returns a stream consisting of the distinct elements (according to Object.equals(Object)) of this stream. For ordered streams, the selection of distinct elements is stable (for duplicated elements, the element appearing first in the encounter order is preserved.) For unordered streams, no stability guarantees are made.

Code:

   Integer[] array = new Integer[]{5, 10, 20, 58, 10};
   Stream.of(array)
         .distinct()
         .forEach(i -> System.out.print(" " + i));

Output:

5,10,20,58

Read More About distinct function

Try this :

Set<Integer> uniqueNumbers = new HashSet<Integer>(Arrays.asList(numbers));

uniqueNumbers will contain only unique values

Try this code.. it will work

package Exercises;
import java.util.Scanner;
public class E5Second
{
    public static void main(String[] args) 
    {
        Scanner In = new Scanner(System.in);
        int [] number = new int [10];
        fillArr(number);

        boolean [] distinct = new boolean [10];

        int count = 0; 
        for (int i = 0; i < number.length; i++) 
        {
            if (isThere(number,i) == false)
            {
                distinct[i] = true;
                count++;
            }
        }
        System.out.println("\nThe number of distinct numbers is  " + count);
        System.out.print("The distinct numbers are: ");
        displayDistinct(number, distinct);
    }
    public static void fillArr(int [] number)
    {
        Scanner In = new Scanner(System.in);
        System.out.print("Enter ten integers ");
        for (int i = 0; i < number.length; i++)
            number[i] = In.nextInt();
    }
    public static boolean isThere(int [] number, int i)
    {
        for (int j = 0; j < i; j++)
            if(number[i] == number[j])
                return true;
        return false;
    }
    public static void  displayDistinct(int [] number, boolean [] distinct)
    {
        for (int i = 0; i < distinct.length; i++)
            if (distinct[i]) 
                System.out.print(number[i] + " ");
    }
}

One possible logic: If you're supposed to only sort out "unique" numbers, then you'll want to test each number as it's entered and added to the first array, and loop through the array and see if it's equal to any of the numbers already there; if not, add it to the "unique" array.

Sets in java doesn't allow duplicates:

    Integer[] array = new Integer[]{5, 10, 20, 58, 10};
    HashSet<Integer> uniques = new HashSet<>(Arrays.asList(array));

That's it.

Something like this should work for you:

     Scanner input = new Scanner(System.in);

     // Create arrays & variables  
     int arrayLength = 10;
     int[] numbers = new int[arrayLength];
     int[] distinctArray = new int[arrayLength];
     int count = 0;
     Set<Integer> set = new HashSet<Integer>();

     System.out.println("Program starting...");
     System.out.print("Please enter in " + numbers.length + " numbers: ");

     for (int i = 0; i < numbers.length; i++) {
         set.add(input.nextInt());
     }

     for(Integer i : set)
     {
         System.out.println("" + i);
     }

This will only add unique values to the set.

int a[] = { 2, 4, 5, 3, 3, 3, 4, 6 };
int flag = 0;
for (int i = 0; i < a.length; i++)
{  
    flag = 0;
    for (int j = i + 1; j < a.length; j++)
    {
        if (a[i] == a[j])
        {
            flag = 1;
        }
    }

    if (flag == 0)
    {
        System.out.println(a[i]);
    }
}

You can use stream() and then use distinct() which will remove all the duplicates value from the passed int array.

Code:

    int numbers [] = {1, 2, 3, 2, 1, 6, 3, 4, 5, 2};
    List<Integer> distinctNumbers = Arrays.stream(numbers).boxed().distinct().sorted().collect(Collectors.toList());
    distinctNumbers.stream().forEach(s-> System.out.println(s));