You can’t have the TreeMap itself sort on the values, since that defies the SortedMap specification:
A
Mapthat further provides a total ordering on its keys.
However, using an external collection, you can always sort Map.entrySet() however you wish, either by keys, values, or even a combination(!!) of the two.
Here’s a generic method that returns a SortedSet of Map.Entry, given a Map whose values are Comparable:
static <K,V extends Comparable<? super V>>
SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
new Comparator<Map.Entry<K,V>>() {
@Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
int res = e1.getValue().compareTo(e2.getValue());
return res != 0 ? res : 1;
}
}
);
sortedEntries.addAll(map.entrySet());
return sortedEntries;
}
Now you can do the following:
Map<String,Integer> map = new TreeMap<String,Integer>();
map.put("A", 3);
map.put("B", 2);
map.put("C", 1);
System.out.println(map);
// prints "{A=3, B=2, C=1}"
System.out.println(entriesSortedByValues(map));
// prints "[C=1, B=2, A=3]"
Note that funky stuff will happen if you try to modify either the SortedSet itself, or the Map.Entry within, because this is no longer a “view” of the original map like entrySet() is.
Generally speaking, the need to sort a map’s entries by its values is atypical.
Note on == for Integer
Your original comparator compares Integer using ==. This is almost always wrong, since == with Integer operands is a reference equality, not value equality.
System.out.println(new Integer(0) == new Integer(0)); // prints "false"!!!