Sort an Array of ArrayList<Integer> in java

Question

I want to sort an Array of ArrayList by the first int of the ArrayLists elements. I have tried to override the compare method of the Comparator class but it throws:

Exception in thread "main" java.lang.NullPointerException
at BikeGA$1.compare(BikeGA.java:515)
at BikeGA$1.compare(BikeGA.java:1)
at java.util.TimSort.countRunAndMakeAscending(TimSort.java:351)
at java.util.TimSort.sort(TimSort.java:230)
at java.util.Arrays.sort(Arrays.java:1438)
at BikeGA.main(BikeGA.java:512)

The code is:

int max_generations = 20;
static ArrayList<Integer>[] population = new ArrayList[max_generations];

Arrays.sort(population, new Comparator<ArrayList<Integer>>(){
        @Override
        public int compare(final ArrayList<Integer> entry1, final ArrayList<Integer> entry2){
            return entry1.get(0).compareTo(entry2.get(0));
        }
    });

Can someone help me? Thanks.

did you initialize all ArrayLists in the array, and added at least an element to each of them? — guido
– guido, Commented Apr 27, 2015 at 7:58
Are you sure that all the ArrayLists in population are initialized? — M. Shaw
– M. Shaw, Commented Apr 27, 2015 at 7:58
Yes, I have printed the array before the sort and it's correctly initialized. — Luca Natali
– Luca Natali, Commented Apr 27, 2015 at 8:03

Eran · Accepted Answer · 2015-04-27 08:15:42Z

2

Your Comparator should handle null or empty ArrayLists in order for it to work with any data you put in your array :

    @Override
    public int compare(final ArrayList<Integer> entry1, final ArrayList<Integer> entry2){
        if (entry1 == null && entry2 == null)
            return 0;
        if (entry1 == null)
            return 1;
        if (entry2 == null)
            return -1;
        if (entry1.isEmpty() && entry2.isEmpty())
            return 0;
        if (entry1.isEmpty())
            return 1;
        if (entry2.isEmpty())
            return -1;
        return entry1.get(0).compareTo(entry2.get(0));
    }

This will put the null elements in the end of the array, and the empty lists before them.

edited Apr 27, 2015 at 8:15

answered Apr 27, 2015 at 8:01

Eran

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1 Comment

guido Over a year ago

mmh don't like the fact that two null elements will be sorted in a different order depending on which one goes in the first argument. if both are null or empty you should return 0

Jordi Castilla · Accepted Answer · 2015-04-27 08:00:18Z

0

You are comparing 2 arrays but you don't check if comparing object exists...

int max_generations = 20;
static ArrayList<Integer>[] population = new ArrayList[max_generations];

Arrays.sort(population, new Comparator<ArrayList<Integer>>(){
        @Override
        public int compare(final ArrayList<Integer> entry1, final ArrayList<Integer> entry2){
            Integer value1 = entry1.get(0) == null : -1 ? entry1.get(0);
            Integer value2 = entry2.get(0) == null : -1 ? entry2.get(0);

            return value1.compareTo(value2);
        }
    });

answered Apr 27, 2015 at 8:00

Jordi Castilla

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Comments

Mr tintin · Accepted Answer · 2021-11-01 15:39:52Z

0

In my case object was:

public class Edge {
    private Integer weight;
    public Edge(Integer weight) {
        this.weight = weight;
    }
    public Integer getWeight() {
        return weight;
    }
}

And I call sort for (List edges = new ArrayList<>();) from other class:

Collections.sort(edges,(edge1,edge2) -> edge1.getWeight().compareTo(edge2.getWeight()));

Here is the result: Exception img

Solution was to initialize value of object:

public class Edge {
    private Integer weight = 0;
    public Edge(Integer weight) {
        this.weight = weight;
    }
    public Integer getWeight() {
        return weight;
    }
}

and it work!!! code 0

answered Nov 1, 2021 at 15:39

Mr tintin

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