Get minvalue of a Map(Key,Double)

You can use the standard Collections#min() for this.

Map<String, Double> map = new HashMap<String, Double>();
map.put("1.1", 1.1);
map.put("0.1", 0.1);
map.put("2.1", 2.1);

Double min = Collections.min(map.values());
System.out.println(min); // 0.1

Update: since you need the key as well, well, I don't see ways in Collections or Google Collections2 API since a Map is not a Collection. The Maps#filterEntries() is also not really useful, since you only know the actual result at end of iteration.

Most straightforward solution would then be this:

Entry<String, Double> min = null;
for (Entry<String, Double> entry : map.entrySet()) {
    if (min == null || min.getValue() > entry.getValue()) {
        min = entry;
    }
}

System.out.println(min.getKey()); // 0.1

(nullcheck on min left aside)

You still can use Collections.min with a custom Comparator to get the Map.Entry with the lower value:

Map<String, Double> map = new HashMap<String, Double>();
map.put("1.1", 1.1);
map.put("0.1", 0.1);
map.put("2.1", 2.1);
Entry<String, Double> min = Collections.min(map.entrySet(), new Comparator<Entry<String, Double>>() {
    public int compare(Entry<String, Double> entry1, Entry<String, Double> entry2) {
        return entry1.getValue().compareTo(entry2.getValue());
    }
});
System.out.printf("%s: %f", min.getKey(), min.getValue()); // 0.1: 0.100000

With Java 8:

Entry<String, Double> min = Collections.min(map.entrySet(),
                                       Comparator.comparing(Entry::getValue));

In traditional way, I would have to sort the map according to the values, and take the first/last one. thanks

No, you wouldn't. You would have to iterate through all values and at each step compare the current element with the smallest one seen so far. That's O(n), compared with O(n*log(n)) for sorting - a potentially huge difference.

BTW, this is exactly how Collections.min() works.