sorting 2 dimensional java array

Question

I have implemented bubble sort to sort a two dimensional java long [][] but my god is it slow, I will be needing the fasted algorithm possible as i will be generating a array of the max heap size jvm will allow me,

So i think the best and fastest way would be to use the inbuild java Arrays.sort

I dont mind if it can only sort on column one as i can change my program to suit, I came across this but am not to familar with comaparator,

this will allow me to sort a dimensional array of integers, does anyone know how to change this to allow longs?, i did thinker around with it with no joy yet.

int d2 [][] = {{1,43},{26,98},{44,398},{11,34},{17,32}};

java.util.Arrays.sort(d2, new java.util.Comparator<int[]>() {
    public int compare(int[] a, int[] b) {
        return b[0] - a[0];
    }
});

i want to sort say

long d2L [][] = {{1,43},{26,98},{44,398},{11,34},{17,32}};

casting is not an option as the numbers a massive

Also if anyone thinks theres a faster method to sort im all ears:)

Answer 1

This sorts based on all columns in O(NlogN), i.e. really fast:

import java.util.*;

class Compare2DArray implements Comparator {
  public int compare(Object a, Object b) {
    int aa[] = (int[]) a;
    int bb[] = (int[]) b;
    for (int i = 0; i < aa.length && i < bb.length; i++)
      if (aa[i] != bb[i])
        return aa[i] - bb[i];
    return aa.length - bb.length;
  }
}

class sort2d {
  public static void main(String args[]) {
    int d2 [][] = {{1,43},{26,98},{44,398},{11,34},{17,32}};
    Arrays.sort(d2, new Compare2DArray());
    for (int i = 0; i < d2.length; i++) {
      for (int j = 0; j < d2[i].length; j++)
        System.out.print(d2[i][j] + " ");
      System.out.println();
    }
  }
}

http://ideone.com/TjEOL

Or you could use generics to avoid casting:

class Compare2DArray implements Comparator<int[]> {
  public int compare(int a[], int b[]) {
    for (int i = 0; i < a.length && i < b.length; i++)
      if (a[i] != b[i])
        return a[i] - b[i];
    return a.length - b.length;
  }
}