CODEWITHSUNDEEP
javaregexstring
daark
daark

Reputation: 326

Splitting alphanumeric strings without a deliminator

I'm trying to implement a comparator for a TreeMap where the key entries are strings of the form

1a
2b
11a
11b
14 
16

I can identify the strings that need extra processing easy enough using this regex

[0-9]+[a-zA-Z]+

using a simple [0-9]+ regex I can find the initial number on the string easily, my question is how can i split these up to let me then compare the integer values and the string characters separately?

Edit: Sample data is above the expected output ideally would be a string array with position 0 being the integer value and position 1 being the String value i.e.

[1,a]
[2,b]
[11,a]
[11,b]

Upvotes: 1

Views: 78

Answers (1)

aioobe
aioobe

Reputation: 421310

Here's one approach that uses the regular expression that you suggest:

new Comparator<String>() {
    Pattern p = Pattern.compile("([0-9]+)[a-zA-Z]+");

    private String getNum(String s) {
        Matcher m = p.matcher(s);
        return m.matches() ? m.group(1) : s;
    }

    @Override
    public int compare(String o1, String o2) {
        o1 = getNum(o1);
        o2 = getNum(o2);
        return Integer.compare(Integer.parseInt(o1),
                               Integer.parseInt(o2));
    }
};

If you're on Java 8, you could do

private static Pattern p = Pattern.compile("([0-9]+)[a-zA-Z]+");
private static int getNum(String s) {
    Matcher m = p.matcher(s);
    return Integer.parseInt(m.matches() ? m.group(1) : s);
}

and then use

Comparator.comparing(YourClass::getNum))

Another approach that doesn't use the regexp that you propose is to do

Comparator.comparing(s -> Integer.parseInt(s.replaceAll("[a-zA-Z]", ""))));

Upvotes: 2

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