String Merge Sort Implementation

Question

I have been trying to implement various types of sort in a program am working on. So far i've managed to sort integers.What changes have to be made in order to make this (merge) code sort a String array instead of an int one? Will the time complexity vary? if so, for better or for worse?

EDIT 1: tried to use the compareTo. something does not seem right. Errors are returned e.g Cannot convert from string to int and vice versa. FIXED

EDIT 2: im getting NullPointerException at line if (array[low].compareTo(array[high]) >= 0) . Suggestions are always welcome.

This is the error:

null  null  null  null  null  
Exception in thread "main" java.lang.NullPointerException
    at Merge.mergeSort_srt(Merge.java:28)
    at Merge.Sort(Merge.java:15)
    at Sort.main(Sort.java:73)

import java.io.File;


public class Merge 
{
    public void Sort (LinkedList listIn, int size) throws Exception
    {
        String[] mergeArray = new String[size] ;
        String textContent = null ;
        File outputFile ;

         for(int i = 0; i < mergeArray.length; i++)
           System.out.print( mergeArray[i]+"  ");
           System.out.println();
           mergeSort_srt(mergeArray,0, mergeArray.length-1);
           System.out.print("Values after the sort:\n");
           for(int i = 0; i <mergeArray.length; i++)
           System.out.print(mergeArray[i]+"  ");
           System.out.println();
           System.out.println("PAUSE");
    }


     public static void mergeSort_srt(String array[],int lo, int n)
     {
           int low = lo;
           int high = n;
           if (array[low].compareToIgnoreCase(array[high]) >= 0)
           {
               return;
           }

           int middle = ((n+1)/ 2);
           mergeSort_srt(array, low, middle);
           mergeSort_srt(array, middle + 1, high);
           int end_low = middle;
           int start_high = middle + 1;
           while ((array[low].compareToIgnoreCase(array[end_low]) <= 0) && (array[start_high].compareToIgnoreCase(array[high]) <= 0))
           {
               if(array[low].compareToIgnoreCase(array[start_high]) < 0)
               {
                   low++;
               }
               else
               {
                   String Temp = array[start_high];
           for (int k = start_high- 1; k >= low; k--)
           {
               array[k+1] = array[k];
           }
           array[low] = Temp;
           low++;
           end_low++;
           start_high++;
           }
           }
     }

}

Don't put the word EDIT in the title if you edit a post. Editing is the norm, not an exception at Stack Overflow. — teukkam
– teukkam, Commented Apr 20, 2012 at 10:42

talnicolas · Accepted Answer · 2012-04-19 16:24:28Z

6

It depends on the way you want to sort your Strings, but the method compareTo of String would help you achieve that kind of sorting:

Returns: the value 0 if the argument string is equal to this string; a value less than 0 if this string is lexicographically less than the string argument; and a value greater than 0 if this string is lexicographically greater than the string argument.

answered Apr 19, 2012 at 16:24

talnicolas

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6 Comments

serge Over a year ago

so instead of the if (array[low] < array[start_high]) i should have array[low].toCompareTo(array[start_high]) ? As for the way of sorting that would be ascending

talnicolas Over a year ago

@voth1234 if(array[low].compareTo(array[start_high]) < 0) will mean that array[low] is inferior to array[start_high]

serge Over a year ago

i see. i will try to do it and if i have touble i will post back. thank you

ratchet freak Over a year ago

@voth1234 actually it would be array[low].compareTo(array[start_high])<0 and he means case sensitive or case insensitive, numerical (as numbers) of lexicographic, the default is case sensitive and lexicographic

talnicolas Over a year ago

@voth1234 then you can see compareToIgnoreCase just behind compareTo in the doc.

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