[Java] Sorting tips when strings and numbers are mixed

Continuing from the previous [Java 8] Proper use of Comparable and Comparator from the viewpoint of employee sorting, these are tips for sorting.

I couldn't find much when I searched for it,

["2", "32", "100", "13", "1", "21"]

Such an array

["A02", "12", "A11", "A01", "B01", "A21"]

I want to sort this array ** "Yoshinani" **! Have you ever thought?

It is troublesome to parse numbers one by one, and when parsed, it is troublesome to control when it is not a number. It's not a big deal, but I thought it could be solved with a relatively simple process, so I'll write down the process.

Expected sort result

["1", "2", "13", "21", "32", "100"]
// ->Before sorting["2", "32", "100", "13", "1", "21"]

["12", "A01", "A02", "A11", "A21", "B01"]
// ->Before sorting["A02", "12", "A11", "A01", "B01", "A21"]

To conclude, ** first compare the lengths of the strings, then compare the strings **.

I will actually write it.

Compare the lengths of the strings, then compare the strings

pubic List<String> sort(List<String> list) {
    list.sort((s1, s2) -> {
        if (compareLength(s1, s2) == 0) {
            return s1.compareTo(s2);
        }
        return compareLength(s1, s2);
    });
    return list;
}

private int compareLength(String s1, String s2) {
    if (s1.length() > s2.length()) {
        return 1;
    } else if (s1.length() < s2.length()) {
        return -1;
    } else {
        return 0;
    }
}

Execution result

List<String> list1 = new ArrayList<>(Arrays.asList("2", "32", "100", "13", "1", "21"));↲
List<String> list2 = new ArrayList<>(Arrays.asList("A02", "12", "A11", "A01", "B01", "A21"));↲
List<String> list3 = new ArrayList<>(Arrays.asList("0002", "0032", "0100", "0013", "0001", "0021"));↲

System.out.println(ArrayUtils.toString(sort(list1)));
System.out.println(ArrayUtils.toString(sort(list2)));

//Execution result
[1, 2, 13, 21, 32, 100]
[12, A01, A02, A11, A21, B01]

By the way, even if the digits are padded with 0s, if the number of digits is the same, it will sort as expected.

When filling with 0

List<String> list = new ArrayList<>(Arrays.asList("0002", "0032", "0100", "0013", "0001", "0021"));↲

System.out.println(ArrayUtils.toString(sort(list)));

//Execution result
[0001, 0002, 0013, 0021, 0032, 0100]

I don't think it can be used in any situation because it says ** "Yoshinani" **, but personally I am satisfied with the results I expected. If you have logic that can be processed smarter, or if you have a list that can not be processed "good" with this method, I would be grateful if you could share it!

Postscript

He commented that it would be simpler if you did the following!

List<String> list = Arrays.asList("2", "32", "100", "13", "1", "21");
list.sort(Comparator.comparing(String::length).thenComparing(Function.identity());
System.out.println(list);