I am bugged !!!

[ajaygargnsit]

Kindly explain why the following mergesort code does not give the desired result? Specifically, the array 'after-it-is-supposed-to-be-sorted' is exactly the same as the original array. Plzzzzzzzzzz help !!!


package src;
import java.util.*;

class Sorti
{
public int a[] = {287,30,58,122,390,4,21,65,2987,459};
public int f = 0;
public int l = 9;

public void display()
{
for(int m = 0; m < a.length; m++)
{
System.out.println(a[m]);
}
}
}


class Mergesort extends Sorti
{
public int mid = 0;
public int i, j, k;
public int[] temp1, temp2;

public void mergesort(int[] arr, int start, int end)
{
mid = (start + end) / 2;
if(start < end)
{
mergesort(arr,start,mid);
mergesort(arr,mid+1,end);
merge(arr,start,mid,end);
}

/*if(end > 0)
{
System.out.println(end);
mergesort(arr,start,end-1);

}*/
}

public void merge(int[] arr, int start, int mid, int end)
{
int tema[] = new int[mid - start + 2];
int temb[] = new int[end - mid + 1];

System.out.println(" First : " + (mid-start+2));
System.out.println("Second : " + (end-mid+1));

for(i = 0; i < (mid - start + 1); i++)
{
tema[i] = arr[start + i];
}
tema[i] = 10000;


for(i = 0; i < (end - mid); i++)
{
temb[i] = arr[mid + 1 + i];
}
temb[i] = 10000;

i = 0;
j = 0;

for(k = 0; k < (end - start + 1); k++)
{
if(tema[i] < temb[j])
{
arr[k + start] = tema[i];
i++;
}
else
{
System.out.println("arrrrrrrr");
arr[k + start] = temb[j];
j++;
}
}
}
}



public class sort
{
public static void main(String[] args)
{
Mergesort obj = new Mergesort();

obj.display();

obj.mergesort(obj.a, obj.f, obj.l);

obj.display();
}
}

[CaptainMorgan]

Is there a reason why you're using the original array as you are and not a random array? Just curious.

I wouldn't have written it as you did.. trying to decipher your code is a little time consuming when you're not used to someone's style of code.. but when I have more time I'd like to see if I can be of further assistance. In the meantime, your code is very similar to that found here, maybe you can figure it out before I write again.