Sort on a string that may contain a number

Question

I need to write a Java Comparator class that compares Strings, however with one twist. If the two strings it is comparing are the same at the beginning and end of the strin

Answer 1

Here is the solution with the following advantages over Alphanum Algorithm:

1. 3.25x times faster (tested on the data from 'Epilogue' chapter of Alphanum description)
2. Does not consume extra memory (no string splitting, no numbers parsing)
3. Processes leading zeros correctly (e.g. "0001" equals "1", "01234" is less than "4567")

public class NumberAwareComparator implements Comparator
{
    @Override
    public int compare(String s1, String s2)
    {
        int len1 = s1.length();
        int len2 = s2.length();
        int i1 = 0;
        int i2 = 0;
        while (true)
        {
            // handle the case when one string is longer than another
            if (i1 == len1)
                return i2 == len2 ? 0 : -1;
            if (i2 == len2)
                return 1;

            char ch1 = s1.charAt(i1);
            char ch2 = s2.charAt(i2);
            if (Character.isDigit(ch1) && Character.isDigit(ch2))
            {
                // skip leading zeros
                while (i1 < len1 && s1.charAt(i1) == '0')
                    i1++;
                while (i2 < len2 && s2.charAt(i2) == '0')
                    i2++;

                // find the ends of the numbers
                int end1 = i1;
                int end2 = i2;
                while (end1 < len1 && Character.isDigit(s1.charAt(end1)))
                    end1++;
                while (end2 < len2 && Character.isDigit(s2.charAt(end2)))
                    end2++;

                int diglen1 = end1 - i1;
                int diglen2 = end2 - i2;

                // if the lengths are different, then the longer number is bigger
                if (diglen1 != diglen2)
                    return diglen1 - diglen2;

                // compare numbers digit by digit
                while (i1 < end1)
                {
                    if (s1.charAt(i1) != s2.charAt(i2))
                        return s1.charAt(i1) - s2.charAt(i2);
                    i1++;
                    i2++;
                }
            }
            else
            {
                // plain characters comparison
                if (ch1 != ch2)
                    return ch1 - ch2;
                i1++;
                i2++;
            }
        }
    }
}