In Java, sort hash map by its key.length()

Question

i have a hashmap like this:

HashMap<String,Integer> map = new HashMap<String,Integer>();
map.put("java",4);
map.put("go",2);
map.put("objective-c",11);
map.put("c#",2);

now i want to sort this map by its key length, if two keys length are equal (e.g go and c# both length 2), then sorted by alphba order. so the outcome i expect to get is something like:

printed result: objective-c, 11 java, 4 c#, 2 go, 2

here is my own attamp, but it doesnt work at all...

      HashMap<String,Integer> map = new HashMap<String,Integer>();
          map.put("java",4);
          map.put("go",2);
          map.put("objective-c",11);
          map.put("c#",2);

      Map<String,Integer> treeMap = new TreeMap<String, Integer>(
                new Comparator<String>() {
                    @Override
                    public int compare(String s1, String s2) {
                        return s1.length().compareTo(s2.length());
                    }
                }
        );

actually the 'compareTo' method appears as red (not be able to compile).... please someone help me with some code example...i am a bit confusing with how to use comparator class to customize compare object...

Answer 1

The compiler is complaining because you cannot call compareTo on an int. The correct way to sort the map is the following:

Map<String, Integer> treeMap = new TreeMap<String, Integer>(
    new Comparator<String>() {
        @Override
        public int compare(String s1, String s2) {
            if (s1.length() > s2.length()) {
                return -1;
            } else if (s1.length() < s2.length()) {
                return 1;
            } else {
                return s1.compareTo(s2);
            }
        }
});

The first two conditions compare the lengths of the two Strings and return a positive or a negative number accordingly. The third condition would compare the Strings lexicographically if their lengths are equal.