(Java) How do I remove duplicates in the calculation of possible sums of numbers?

Question

import java.util.ArrayList;

public class Testing {
    public static void main(String[] args) {
        ArrayList<Double> valid = new ArrayList<Double>();
        ArrayList<Double> result = new ArrayList<Double>();
        valid.add(0.05);
        valid.add(0.05);
        valid.add(0.1);
        getSum(valid, result, 0, 0);
        System.out.println(result);
    }

    public static void getSum(ArrayList<Double> valid, ArrayList<Double> result, int start, double sum) {
        if (valid.size() == start) {
            result.add(sum);
            return;
        }

        double value = sum + valid.get(start);

        getSum(valid, result, start + 1, value);
        getSum(valid, result, start + 1, sum);
    }
}

I want to not even enter the duplicates into the result arrayList, as removing duplicates after will result in not enough memory, as the result arrayList gets too big too fast. How do I do this?

Edit: A hashset suffices.

Start here: https://stackoverflow.com/questions/1035008/what-is-the-difference-between-set-and-list — Kon, Apr 04 '18 at 19:12
sure. If you run into other issues feel free to come back - make sure to detail what you tried, why it failed, and what you expected that didn't occur. — Kon, Apr 04 '18 at 19:15

score 0 · Accepted Answer · answered Apr 04 '18 at 19:22

0

Using interfaces gives greater code universality, in java 8, you can use stream and method distinct():

List<Double> valid = new ArrayList<>();
valid.add(0.05);
valid.add(0.05);
valid.add(0.1);
List<Double> removedDuplicates = valid.stream().distinct().collect(Collectors.toList());

answered Apr 04 '18 at 19:22

Kamil Tomasz Jarmusik

140
8

Using a set is much easier. – Raymo111 Apr 04 '18 at 19:25
If at the entrance and exit the List needs to be or elements must be in a fixed order, it is an alternative. I do not mean this example. – Kamil Tomasz Jarmusik Apr 04 '18 at 19:30

score 0 · Answer 2 · answered Apr 04 '18 at 19:23

I ended up using a set instead of an ArrayList like Kon and Sam_ferree suggested. This ended up working:

import java.util.ArrayList;
import java.util.Set;
import java.util.TreeSet;

public class Testing {
    public static void main(String[] args) {
        ArrayList<Double> valid = new ArrayList<Double>();
        Set<Double> result = new TreeSet<Double>();
        valid.add(0.05);
        valid.add(0.05);
        getSum(valid, result, 0, 0);
        System.out.println(result);
    }

    public static void getSum(ArrayList<Double> valid, Set<Double> result, int start, double sum) {
        if (valid.size() == start) {
            result.add(sum);
            return;
        }

        double value = sum + valid.get(start);

        getSum(valid, result, start + 1, value);
        getSum(valid, result, start + 1, sum);
    }
}

score -1 · Answer 3 · answered Apr 04 '18 at 19:11

-1

Use a Set instead of a List, like HashSet or TreeSet

Sets by definition don't allow duplicates, so changing that datatype is the only change your code needs.

answered Apr 04 '18 at 19:11

sam_ferree

57
5

(Java) How do I remove duplicates in the calculation of possible sums of numbers?

3 Answers3