Sorting a String Array using a Comparator based on a Character in the String

Question

So I'm trying to sort a String Array using a Comparator but the sorting is based on the length of String then on the third Character of the String.

This is my Comparator so far:

class StringSorter implements Comparator<String> {
    public int compare(String s1, String s2) {
        if(s1.length() < s2.length()) {
            return -1;
        }
        if(s1.length() > s2.length()) {
            return 1;
        }    
        return s1.charAt(3)+"".compareTo(s2.charAt(3)+"");
    }
}//Comparator

This line return s1.charAt(3)+"".compareTo(s2.charAt(3)+""); is my attempt so far and I get an IndexOutOfBoundsException but every String in my Array has at least the length of 4 so I don't understand why the error.

As for my question, why do I get that error? and is that how I should write my Comparator if I'm to sort based on length and a character in the String?

Edit: The Arrays I need to process follows this format

{"1:bbbbb", "2:aaa", "=:ccc", "1:qqqq", "1:eeee", "=:zzz", "1:vvv", "2:oooo", "=:eee", "1:fffff"}

you question needs to be verifiable, so please provide sample inputs — alirabiee
– alirabiee, Commented Sep 1, 2017 at 15:59
"but every String in my Array has at least the length of 4 so I don't understand why the error." It is wrong as otherwise you would not have the exception. Set a breakpoint and run in debug mode or other way : add a check and throw an exception if the String length is superior to 4. — davidxxx
– davidxxx, Commented Sep 1, 2017 at 15:59
Observation : does compareTo get called on "" or on s1.charAt(3)+""? Interesting order precedence thing here, maybe. — Karl Reid
– Karl Reid, Commented Sep 1, 2017 at 16:00
Note that your code could be simplified (and made correct, and made clearer, and made faster) to Comparator<String> c = Comparator.comparingInt(String::length).thenComparingInt(s -> s.charAt(3));. — JB Nizet
– JB Nizet, Commented Sep 1, 2017 at 16:02
No need to do all to convert to string etc. Just use- new Character(s1.charAt(2)).compareTo(s2.charAt(2)) — nagendra547
– nagendra547, Commented Sep 1, 2017 at 17:10

Community · Accepted Answer · 2020-06-20 09:12:55Z

Your code works fine. I've tweaked it a bit here to show you how you can make it more resilient though.

class StringSorter implements Comparator<String> {
    private final int pos;

    public StringSorter(int pos) {
        this.pos = pos;
    }

    public int compare(String s1, String s2) {
        if (s1.length() < s2.length()) {
            return -1;
        }
        if (s1.length() > s2.length()) {
            return 1;
        }
        if ( s1.length() <= pos ) {
            return s1.compareTo(s2);
        }
        return Character.compare(s1.charAt(pos),s2.charAt(pos));
    }
}

public void test() {
    String[] test2 = {"1:bbbbb", "2:aaa", "=:ccc", "1:qqqq", "1:eeee", "=:zzz", "1:vvv", "2:oooo", "=:eee", "1:fffff"};
    System.out.println("Before: " + Arrays.toString(test2));
    Arrays.sort(test2, new StringSorter (2));
    System.out.println("After:  " + Arrays.toString(test2));
}

prints:

Before: [1:bbbbb, 2:aaa, =:ccc, 1:qqqq, 1:eeee, =:zzz, 1:vvv, 2:oooo, =:eee, 1:fffff]

After: [2:aaa, =:ccc, =:eee, 1:vvv, =:zzz, 1:eeee, 2:oooo, 1:qqqq, 1:bbbbb, 1:fffff]

I suspect you are using 3 meaning the 3rd character - this is wrong. The 3rd character is at position 2;

Remember - the nth character in a String is indexed by [n-1] because the first is at 0.

nagendra547 · Accepted Answer · 2017-09-01 17:08:09Z

1

This is solution for your problem using ternary operator.

Simple one liner comparison method.

class StringSorter implements Comparator<String> {

    @Override
    public int compare(String s1, String s2) {
        return s1.length() - s2.length() != 0 ? s1.length() - s2.length() :  new Character(s1.charAt(2)).compareTo(s2.charAt(2));
    }
}

answered Sep 1, 2017 at 17:08

nagendra547

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