How to get values of an array in the alphabetical order?

Question

I have an array of strings plus one additional string. I want to use this string and values of array to get a set of string. Then I want to order the set of string alphabetically and extract the string which is the first in the list. What is the easiest way to do it in Java?

ADDED:

I wanted to do it this way:

List<String> playersList = Arrays.asList(players);
playersList.add(userName); // <---------- HERE IS A PROBLEM
Collections.sort(playersList);

I do not get any errors during the compilation. But during the execution I get a "UnsopportedOperationException". And it happens in the second line.

Answer 1

Either append the value to the array and sort it with Arrays.sort or create a List of the items and sort them using Collections.sort. The natural ordering of the strings will be alphabetical.

Answer 2

If you just want to get the minimum of an array of String with an additional external element, then you don't have to sort and extract first (which would be O(N log N)). You can do it in O(N).

String minPlayer = Collections.min(Arrays.asList(players));
minPlayer = Collections.min(Arrays.asList(minPlayer, extraPlayer));  

Answer 3

Arrays.asList wraps the array with an unmodifiable List, so when you try to add to it it throws UnsupportedOperationException. What you could do it create a new a ArrayList and add your elements to it, then you're free to modify it afterwards.

List<String> list = new ArrayList<String>(Arrays.asList(players));
list.add(userName);