How to sort Alphanumeric String

Here is a self-contained example on how to do this (not particularly optimized):

final Pattern p = Pattern.compile("^\\d+");
String[] examples = { 
   "1some", "2some", "20some", "21some", "3some", "some", "1abc", "abc"
};
Comparator<String> c = new Comparator<String>() {
    @Override
    public int compare(String object1, String object2) {
        Matcher m = p.matcher(object1);
        Integer number1 = null;
        if (!m.find()) {
            return object1.compareTo(object2);
        }
        else {
            Integer number2 = null;
            number1 = Integer.parseInt(m.group());
            m = p.matcher(object2);
            if (!m.find()) {
                return object1.compareTo(object2);
            }
            else {
                number2 = Integer.parseInt(m.group());
                int comparison = number1.compareTo(number2);
                if (comparison != 0) {
                    return comparison;
                }
                else {
                    return object1.compareTo(object2);
                }
            }
        }
    }
};
List<String> examplesList = new ArrayList<String>(Arrays.asList(examples));
Collections.sort(examplesList, c);
System.out.println(examplesList);

Output

[1abc, 1some, 2some, 3some, 20some, 21some, abc, some]

Explanation

• The example uses a constant Pattern to infer whether a number is in the String‘s starting position.
• If not present in the first String, it compares it as is to the second.
• If present indeed in the first, it checks the second.
• If not present in the second, it compares the two Strings as is, again
• If present in both, it compares the Integers instead of the whole Strings, hence resulting in a numerical comparison rather than a lexicographical one
• If the number compare identical, it goes back to lexicographic comparison of the whole Strings (thanks MihaiC for spotting this one)