CODEWITHSUNDEEP
javaarrayssorting
user2821242
user2821242

Reputation: 1181

Sort array of strings lexicographically based on last alphabet - String index out of range -1 Exception

package mediumchallenge;

import java.util.Arrays;
import java.util.Comparator;

public class sortarrayofstringsbylength{

    private static class comparelengths implements Comparator<String>{

        @Override
        public int compare(String arg0, String arg1) {
            // TODO Auto-generated method stub
            return (arg0.length()- arg1.length());
        }
    }

    private static class lastchar implements Comparator<String>{

        @Override
        public int compare(String s1, String s2) {
            // TODO Auto-generated method stub
            return (s1.charAt(s1.length()-1) - s2.charAt(s2.length()-1));
        }
    }

    public static void main(String args[]){
        String arr[] = {"abd" , "abc","y" , "rtyuodfdjfhd","weq","wge",""};
        Arrays.sort(arr);
        for(int i=0;i<arr.length;i++){
            System.out.println(arr[i]);
        }

        // sorts according to length of string  
        Arrays.sort(arr, new comparelengths());
        for(int j=0;j<arr.length;j++){
            System.out.println(arr[j]);
        }
        //sorts lexicographically according to last letter  
        Arrays.sort(arr, new lastchar());
        for(int k=0;k<arr.length;k++){
            System.out.println(arr[k]);
        }
    }
}

I'm trying to sort the string array based on last character of each string using comparator. It's giving me String index out of range -1 exception

Upvotes: 0

Views: 1134

Answers (2)

user3214409
user3214409

Reputation: 1

String arr[] = {"abd" , "abc","y" , "rtyuodfdjfhd","weq","wge",""};

The last element is "", it's length is zero. The code "s1.charAt(s1.length()-1)" will throw String index out of range -1 exception.

Upvotes: 0

Menelaos
Menelaos

Reputation: 26279

A solution would be to check the lengths themselves for zero.

  • if s2 is empty and s1 is not empty, s2 should go behind s1 so we return -1.
  • if s1 is empty and s2 is not empty, s1 should go behind s2 so we return 1.
  • if s2 is empty and s1 is empty, they are equal and we should return 0.

Below an example of such an implementation:

@Override
public int compare(String s1, String s2) {

// Returns a negative integer, zero, or a positive integer 
// as the first argument is less than, equal to, or greater than the second.

                if(s1.length() == 0 && s2.length() > 0)return 1; 
                if(s2.length() == 0 && s1.length() > 0)return -1; 
                if(s2.length() == 0 && s1.length() == 0)return 0; 

                return (s1.charAt(s1.length()-1) - s2.charAt(s2.length()-1));
            }

Basically you need to decide if you will allow empty strings and what action to take when they are encountered (e.g. push them below).

If you want empty strings to be first use:

if(s1.length() == 0 && s2.length() > 0)return -1; 
if(s2.length() == 0 && s1.length() > 0)return 1; 
if(s2.length() == 0 && s1.length() == 0)return 0;

If you want empty strings to be last use:

if(s1.length() == 0 && s2.length() > 0)return 1; 
if(s2.length() == 0 && s1.length() > 0)return -1; 
if(s2.length() == 0 && s1.length() == 0)return 0;

Upvotes: 1

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