Checking for duplicate values in arrayList

Question

I'm not sure how I should check for duplicates in my arrayList. I want to return true or false in my output depending on whether or not the array contains duplicate numbers. The numbers are being pulled from another class.

public class Data {

    private ArrayList<Double> sets;

    public Data(double[] set) {
        this.sets = new ArrayList<Double>();
        for (double i : set) {
            this.sets.add(i);
        }
    }       
    public double hasDuplicate(){
        for (int i = 0; i < 6; i++) {
               ArrayList<Double> sets = new ArrayList<Double>();
               for (int j = 0; j < 6; j++)
                sets.add(); //confused with this
               }

    }

Why did you instantiate a new ArrayList in your loop ? you want to check if it has duplicate right? — A Sdi
– A Sdi, Commented Dec 13, 2016 at 19:25
Why your hasDuplicate method returns double and not boolean? Why are you iterating up to 6 only? You could just throw in your numbers into the set and then check whether size of the set is same as size of the list: return new HashSet(sets).size() != sets.size() — Jaroslaw Pawlak
– Jaroslaw Pawlak, Commented Dec 13, 2016 at 19:39

Steven Waterman · Accepted Answer · 2016-12-13 19:32:54Z

1

Iterate over the elements, keeping track of which you've seen. If you see one you've already seen, you can return true.

public boolean hasDuplicate(){
    Set<Double> seenValues = new HashSet();
    for(Double value : sets){
        if(seenValues.contains(value){
            return true;
        }
        else{
            seenValues.add(value);
        }
    }
    return false;
}

answered Dec 13, 2016 at 19:32

Steven Waterman

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Comments

robot_alien · Accepted Answer · 2016-12-13 19:41:34Z

0

You are creating a new Arraylist of sets 6 times in your code. Which is unnecessarily a waste. Initialize it once. Keep adding values to it in the for loop. Also use List sets = new ArrayList(); By this it becomes implementation independent. Like in future you could use LinkedList(); for the same without much change.

answered Dec 13, 2016 at 19:41

robot_alien

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