Selection Sort using a comparable class

Question

I made java Selection Sort using Comparable class, File scanner.

In this code, we get txt file's name and store all words in String[] list and show index and stored word.

Finally, we sort this String[] list using Selection Sorting and check how much time was spent. but there's some error code.

This is an AbstractSort class

abstract class AbstractSort
{
    public static void sort(Comparable[] a) { };

    protected static boolean less(Comparable v, Comparable w )
    {
        return v.compareTo(w) < 0; // This Line is Error
    }

    protected static void exch(Comparable[] a, int i, int j)
    {
        Comparable t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    protected static void show(Comparable[] a)
    {
        for(int i = 0; i < a.length; i++)
            System.out.println(a[i] + " ");
        System.out.println();
    }

    protected static boolean isSorted(Comparable[] a)
    {
        for(int i = 1; i < a.length; i++)
        {
            if(less(a[i], a[i - 1])) // This Line is also Error
                return false;
        }
        return true;
    }
}

and this is a Selection Sort class which is extends AbstractSort class

class Selection extends AbstractSort {
    public static void sort(Comparable[] a) {
        int n = a.length;

        for(int i = 0; i < n - 1; i++) {
            int min = i;

            for(int j = i + 1; j < n; j++) {
                if(less(a[j], a[min]))
                    min = j;
            }
            exch(a, i, min);
        }
        assert isSorted(a);
    };
}

and this is main function

public class HW1{
    static String[] resize(int idx, String[] arr) {
        String[] temp = new String[idx * 2];

        for(int i = 0; i < idx; i++)
            temp[i] = arr[i];

        return temp;
    }

    public static void main(String args[]) {
        int INIT_LEN = 10000;
        long start, end, time;
        String[] list = new String[INIT_LEN];
        Scanner sc = new Scanner(System.in);
        int idx = 0;

        try {
            System.out.println("File Name?");
            String src = sc.nextLine();
            sc = new Scanner(new FileInputStream(src));

            while(sc.hasNext()) {
                String word = sc.next().toString();

                if(idx == list.length)
                    list = resize(idx, list);

                list[idx++] = word;
            }
            System.out.println("1. Total Word = " + idx);
            for(int i = 0; i < idx; i++)
                System.out.println(i + "idx:" + list[i]);

            start = System.currentTimeMillis();
            Selection.sort(list); // This Line is also Error            
            end = System.currentTimeMillis();
            time = end - start;
            System.out.println("2. Selection Sorted? = true, Time = " + time + "ms");
        }catch (FileNotFoundException fnfe) {
            System.out.println("No File");
        }catch (IOException ioe) {
            System.out.println("Can't Read File");
        }
    }
}

when I run this code, I can see all words are stored int String[] list but there's also error code together.

Exception in thread "main" java.lang.NullPointerException
at AbstractSort.less(HW1.java:8)
at Selection.sort(HW1.java:40)
at HW1.main(HW1.java:84)

I don't know why this error code is occured...

score 1 · Answer 1 · answered Mar 18 '19 at 13:47

When you call Selection.sort(list) in main, it appears that the list has a length of 10000.

Every element defaults to null.

So if you read in three words your list will look like this:

word1,word2,word3,null,null,null......null

Quick hack so you don't need to resize the array - try making your inner loop in Selection::sort:

    for (int j = i + 1; j < n; j++) {
        if (a[j] == null) {
            break;
        }

        if (less(a[j], a[min]))
            min = j;
    }

Or - resize the array appropriately before processing.

Or - use an ArrayList to push words to and then convert to an array if you absolutely must use an array.

Selection Sort using a comparable class

1 Answers1