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  • Sort alphanumeric string with awareness of number using Java comparator

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    The java by default sorting is based on either number or string. If we try to sort a alphaneumeric it will be treated as an string so the sorting will come something like: 1abc 11abc 2abc

    instead of 1abc 2abc 11abc

    So in such cases we can use our own comparator based on regex, code is given below:

    import java.math.BigInteger;
    import java.util.Comparator;
    import java.util.regex.Matcher;
    import java.util.regex.Pattern;
    
    public class CompareNumberBasedString implements Comparator<CharSequence> {
    
        private static final Pattern PATTERN = Pattern.compile("(\\D*)(\\d*)");
    
        public int compare(CharSequence c1, CharSequence c2) {
            Matcher matcher1 = PATTERN.matcher(c1);
            Matcher matcher2 = PATTERN.matcher(c2);
    
            // The only way find() could fail is at the end of a string
            while (matcher1.find() && matcher2.find()) {
    
                //non digit comparison
                int nonDigitCompare = matcher1.group(1).compareTo(matcher2.group(1));
                if (0 != nonDigitCompare) {
                    return nonDigitCompare;
                }
    
                // digit comparison
                if (matcher1.group(2).isEmpty()) {
                    return matcher2.group(2).isEmpty() ? 0 : -1;
                } else if (matcher2.group(2).isEmpty()) {
                    return +1;
                }
    
                BigInteger number1 = new BigInteger(matcher1.group(2));
                BigInteger number2 = new BigInteger(matcher2.group(2));
                int numberCompare = number1.compareTo(number2);
                if (0 != numberCompare) {
                    return numberCompare;
                }
            }
    
            // Handle if one string is a prefix of the other.
            return matcher1.hitEnd() && matcher2.hitEnd() ? 0 :
                matcher1.hitEnd()                ? -1 : +1;
        }
    }
    

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