delete repeated values array java

Following is a direct implementation, without using the Collections API, which results in an unrepeatable vector:

import java.util.Arrays;

public class TestIntArray {

    public static void main( String[ ] args ) {
        int[ ] original = { 1 , 3 , 5 , 7 , 9 , 5 , 3 };

        // remover repetidos
        int[ ] unicos = new int[ original.length ];
        int qtd = 0;
        for( int i = 0 ; i < original.length ; i++ ) {
            boolean existe = false;
            for( int j = 0 ; j < qtd ; j++ ) {
                if( unicos[ j ] == original[ i ] ) {
                    existe = true;
                    break;
                }
            }
            if( !existe ) {
                unicos[ qtd++ ] = original[ i ];
            }
        }

        // ajuste do tamanho do vetor resultante
        unicos = Arrays.copyOf( unicos , qtd );

        // imprime resultado
        for( int i = 0 ; i < unicos.length ; i++ ) {
            System.out.println( "" + i + " = " + unicos[ i ] );
        }

    }
}

You can adapt this in a method that receives a int[] and returns a int[] .

You can add all in a Set and then print the object:

int[] ja={1,2,3,3,4,5,6,6,8,9};
Set<Integer> set = new HashSet<>();
for(int a: ja) {
    set.add(a);
}
System.out.println(set);

The above code prints:

  

[1, 2, 3, 4, 5, 6, 8, 9]

The Set is a collection of unique objects, so when you add repeated elements to it, those elements are automatically dropped.

Alternative

An alternative to make your code simpler is to use a Integer vector instead of a int vector, so that you can add all elements of the vector to your set variable with just one command, your code would look like this:

Integer[] ja={1,2,3,3,4,5,6,6,8,9};
Set<Integer> set = new HashSet<>();
set.addAll(Arrays.asList(ja));
System.out.println(set);

The result is the same.

Returning to vector

To return the value of Integer[] to vector just do:

Integer[] jb = set.toArray(new Integer[set.size()]);

If you need to transform to vector of set it will have to be something more manual, like:

int cont = 0;
int[] jaresul = new int[ja.length];
for (Integer i : jb) {
    jaresul[cont++] = i;
}

Your int will contain the elements of jaresul without duplicates.

References: Set (Java Platform SE7) ; Arrays.asList (Java Platform SE7)

With Java8:

Integer[] ja = {1,2,3,3,4,5,6,6,8,9};
List<Integer> distinctList = Arrays.asList(ja).stream().distinct().collect(Collectors.toList());

With While to search and For to print ...

        int[] ja = {1, 2, 3, 3, 4, 5, 6, 6, 8, 9,9,7};
        Arrays.sort(ja);
        int[] jj = new int[ja.length];
        int i = 0;
        int j = 0;     
        int x = 0;
        boolean find;
        while (i < ja.length) { 
            if (i == 0){
                jj[j] = ja[i];
                j++;
            } else {
                x = 0;
                find = false;
                while (x < j && find == false){
                    if (jj[x] == ja[i]) { find = true; }                    
                    x++;
                }
                if (find == false){
                    jj[j] = ja[i];
                    j++;
                }
            }
            i++;
        }            
        ja = Arrays.copyOf(jj, j);            
        for(int r : ja){
            System.out.println(r);
        }

This is your solution.

Greetings from Mexico.

int[] ja = {1, 2, 3, 3, 4, 5, 6, 6, 8, 9};
int[] res = Arrays.stream(ja).distinct().toArray();
for (int i = 0; i < res.length; i++) {
    System.out.print(res[i]);
}