Store a graph in ArrayList

Question

I would like to read a graph from the input file and store it in list of arraylist.First list of the input file contains no.of vertices followed by number of edges. then the next line contains edges of the graph. I tried the following code and i'm getting error which is added at the end

import java.io.File;
import java.util.Scanner;
import java.util.ArrayList;
import java.util.List;
public class ReadGraph {

    public static void main(String[] args) {

        try {
            int vertices,edges;
            Scanner input = new Scanner(new File("in.txt"));
            vertices = input.nextInt();
            edges = input.nextInt();
            List<Integer>[] tree = new List[vertices];
            for (int i = 0; i < edges; i++) {
              tree[input.nextInt()].add(input.nextInt());
            }
            System.out.println(vertices);
            System.out.println(edges);
            input.close();
            System.out.println(tree);

        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }

}

Error

java.lang.ArrayIndexOutOfBoundsException: 2
at ReadGraph.main(ReadGraph.java:30)

This is not how you declare a list, you are creating an array of list, which are all nulls hence the NullPointerException. — Jean-François Savard, Jun 28 '16 at 03:23
List tree = new ArrayList<>(); ... Then in your loop simply write tree.add(input.nextInt()); — Jean-François Savard, Jun 28 '16 at 03:25
System.out.println(tree); It will call toString on the list instance, which is well formatted. Or if you want them one by one : tree.forEach(System.out::println); — Jean-François Savard, Jun 28 '16 at 03:27
Note that the result will probably not be what you expect, but that will fix the NPE. Not sure what you are trying to achieve here. — Jean-François Savard, Jun 28 '16 at 03:28
Then this is not a List, this is a list of list. Or a Map of List. — Jean-François Savard, Jun 28 '16 at 03:32
Change your code to Map> tree = new HashMap<>(); ... Then you can do tree.put(input.nextInt(), yourListOfInts); Use a TreeMap if you want them ordered (you named your variable tree so I'm suspicious). — Jean-François Savard, Jun 28 '16 at 03:35
Then List> = new ArrayList<>(); ... But you won't be able to add at random index (input.nextInt) using a list of list. — Jean-François Savard, Jun 28 '16 at 03:37

Store a graph in ArrayList

0 Answers0