From:

Piotr Kobzda <pikob@gazeta.pl>

Newsgroups:

comp.lang.java.help

Date:

Tue, 07 Jul 2009 22:57:52 +0200

Message-ID:

<h30csi$h3u$1@inews.gazeta.pl>

- If you gather all the Point/double pairs and then want to process

them in sorted order, begin by putting them in a List without

regard to order. When you've gathered them all, sort the List

with Collections.sort(), and then traverse the sorted List.

them in sorted order, begin by putting them in a List without

regard to order. When you've gathered them all, sort the List

with Collections.sort(), and then traverse the sorted List.

Will be n log(n) for sorting, but again does not allow for updates.

Some Lists does.

Example:

List<String> l = new ArrayList<String>(Arrays.asList("b", "a", "c"));

Collections.sort(l);

String s = "abc";

int i = Collections.binarySearch(l, s);

l.add(i < 0 ? -(i + 1) : i, s);

This is at par with the SortedSet solution. Inserting n times, each in

time log(n) will also require time n log(n).

time log(n) will also require time n log(n).

Not exactly, SortedSet typically forms the binary tree, so time needed

will typically be O(n^2).

However, the SortedSet

allows for updates.

allows for updates.

Lists allows for updates as well. Typically, with log(n) for lookup,

and amortized constant time for insert (i.e. adding n elements requires

O(n) time). No important difference asymptotically.

So, one sort of List with n elements (usually marge-sort for

RandomAccess lists) with relatively big n is typically faster than n

add()s to SortedSet (tree-sort equivalent). So, I would prefer sort of

unordered List, before using SortedList in this particular scenario.

piotr

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