Vector Comparison Problem - Java

Question

I have a problem with a simple code in Java, I just started learning the language.

The exercise consists of comparing the values of the vector to see if there are repeated elements.

The problem is, it always gives me 0 in output!

I don’t know if it has to do with the cycle while() or by increasing!

Could you give me a hint ? The IDE I use is Netbeans 8.1.

Below is the extract of the code responsible for that, I have a System.out.print, to print the count of the vector resulting from the junction of the listaA and listaB:

    int[] listaA = {2, -5, -121, 102, -35, -2, 0, -125, 802, -10};
    int[] listaB = {6, 99, -1, 12, 1, -2};
    int[] novoVetor;

    novoVetor = new int[listaA.length + listaB.length];  // tamanha do vetor A e o tamanho do vetor B
    int nr_vetorB = listaA.length, rep=0, g=0;

    for (int i = 0; i < listaA.length; i++) {
        novoVetor[i] = listaA[i];
    }
    for (int j = 0; j < listaB.length; j++) {
        novoVetor[nr_vetorB] = listaB[j];
        nr_vetorB++;
    }


    for (int x = 0; x < novoVetor.length; x++) {
        while( g < novoVetor.length) {
            for (int y = 1; y < novoVetor.length; y++) {
                if (novoVetor[x] == novoVetor[y]) {
                    g = novoVetor.length;
                    rep++;
                } else {
                    g++;
                }


            }

        }
        System.out.print(" " +novoVetor[x]);    
        }

        System.out.print("\nElementos repetidos:" +rep);

Regards

Answer 1

You could do it differently:

- Work only with the two arrays being compared, thus avoiding the creation of the third array and the two boot loops;

- In its search structure, it could remove the while (it seems that it does not fit very well according to the purpose), keep only the two "for" where one scan lists A and the other list B, all from position 0, as it is possible to compare them separately;

- In the case of comparison, it would not need an Else, because if the elements are different, there is nothing to do and the search continues freely; if they are equal, then Voce can store both the positions of the arrays that are repeated (create the variables before entering the search) how much to increase the "rep";

- at the end, Voce will have the last number that was repeated (accessing listA or listB by the saved position), and how many times there were repetitions.

Answer 2

Try it like this:

    int[] listaA = {2, -5, -121, 102, -35, -2, 0, -125, 802, -10};
    int[] listaB = {6, 99, -1, 12, 1, -2};
    int[] todos = new int[listaA.length+listaB.length];

    int pt = 0;
    for(int i : listaA){
        todos[pt] = i;
        pt++;
    }
    for(int i : listaB){
        todos[pt] = i;
        pt++;
    }

    int total = 0;
    for(int ptA=0; ptA < todos.length; ptA++){
        for(int ptB=0; ptB < todos.length; ptB++){
            if(ptA == ptB){ // se os ponteiros forem iguais desconsidera..
                continue;
            }
            if(todos[ptA] == todos[ptB]){
                total++;
            }
        }
    }

    total = (total/2); /// pois irá comparar o mesmo item 2 vezes..
    System.out.println("Elementos repetidos: "+total);