Sorting Strings that contains number in Java [duplicate]

by

Try this comparator, which removes all non-digit characters then compares the remaining characters as numbers:

Collections.sort(strings, new Comparator<String>() {
    public int compare(String o1, String o2) {
        return extractInt(o1) - extractInt(o2);
    }

    int extractInt(String s) {
        String num = s.replaceAll("\\D", "");
        // return 0 if no digits found
        return num.isEmpty() ? 0 : Integer.parseInt(num);
    }
});

Here’s a test:

public static void main(String[] args) throws IOException {
    List<String> strings = Arrays.asList("room1.2", "foo1.1", "foo", "room2.3", "room100.999", "room10", "room.3");

    Collections.sort(strings, new Comparator<String>() {
        public int compare(String o1, String o2) {
            return extractInt(o1) - extractInt(o2);
        }

        int extractInt(String s) {
            String num = s.replaceAll("\\D", "");
            // return 0 if no digits found
            return num.isEmpty() ? 0 : Integer.parseInt(num);
        }
    });
    System.out.println(strings);
}

Output:

[foo, room1, room2, room10, room100]

When the numbers are decimals (also demonstrating Java 8+ style):

public static void main(String[] args) {
    List<String> strings = Arrays.asList("room1.2", "foo1.1", "room2.3", "room100.999", "room10", "room.3");
    Collections.sort(strings, Comparator.comparing(Application::extractDouble));
    System.out.println(strings);
}

static double extractDouble(String s) {
    String num = s.replaceAll("[^\\d.]", "");
    // return 0 if no digits found
    return num.isEmpty() ? 0 : Double.parseDouble(num);
}

Result:

[foo, room.3, foo1.1, room1.2, room2.3, room10, room100.999]

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