Re: A suitable container that can sort -- please help!

Piotr Kobzda <>
Tue, 07 Jul 2009 22:57:52 +0200
Simon wrote:

    - 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.

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

Some Lists does.


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

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).

Not exactly, SortedSet typically forms the binary tree, so time needed
will typically be O(n^2).

However, the SortedSet
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.


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