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From: Tom Anderson on 1 Mar 2010 18:24
On Mon, 1 Mar 2010, laredotornado wrote:

> I"m using Java 1.5. Given a java.util.List that I know to have at least
> one element, what is the best way to check that all elements in the list
> are unique ?

List<T> l;
boolean allUnique = l.size() == new HashSet<T>(l).size();

tom

--
Virtually everything you touch has been mined. -- Prof Keith Atkinson
From: Mike Schilling on 1 Mar 2010 18:30
John B. Matthews wrote:
> In article
> <95bd0b1b-e372-4981-a6cf-eed5a58e4461(a)u19g2000prh.googlegroups.com>,
> laredotornado <laredotornado(a)zipmail.com> wrote:
>
>> I'm using Java 1.5. Given a java.util.List that I know to have at
>> least one element, what is the best way to check that all elements in
>> the list are unique ?
>
> You should be able to construct a Set, say TreeSet, of the List
> elements and see if the sizes match.

TreeSet requires the elements be comparable. HashSet will work for any
element type.


From: Tom Anderson on 1 Mar 2010 18:36
On Mon, 1 Mar 2010, John B. Matthews wrote:

> In article
> <95bd0b1b-e372-4981-a6cf-eed5a58e4461(a)u19g2000prh.googlegroups.com>,
> laredotornado <laredotornado(a)zipmail.com> wrote:
>
>> I'm using Java 1.5. Given a java.util.List that I know to have at
>> least one element, what is the best way to check that all elements in
>> the list are unique ?
>
> You should be able to construct a Set, say TreeSet, of the List elements
> and see if the sizes match.

I am tickled that we've all thought of exactly the same solution! Are
there any other interesting ways to do this?

The only improvement, of sorts, i'd offer is:

boolean areAllUnique(List<T> l) {
Set<T> seen = new HashSet<T>();
for (T o: l) if (!seen.add(0)) return false;
return true;
}

Which doesn't need to examine any more elements of the list than it
absolutely has to to decide if the list is all unique. Except for zero-
and one-element lists, that is.

The way i'd do this in shell script is (not tested!):

l="space separated list"
[[ $(echo $l | wc -w) -eq $(echo $l | tr ' ' '\n' | sort -u | wc -w) ]]

Which suggests the following java:

boolean areAllUnique(List<T> l) {
l = new ArrayList<T>(l);
Collections.sort(l);
T prev = null;
for (T cur: l) {
if (cur.equals(prev)) return false;
prev = cur;
}
return true;
}

Which is pretty different, but still built around the idea of sorting and
then detecting duplicates by comparing adjacent elements, as sort -u does.

tom

--
Virtually everything you touch has been mined. -- Prof Keith Atkinson
From: Joshua Cranmer on 1 Mar 2010 18:53
On 03/01/2010 06:36 PM, Tom Anderson wrote:
> if (cur.equals(prev)) return false;

What if there are null entries in the list?

This snippet also, of course, assumes that T is a comparable object (I
believe Collections.sort fails if there's no comparator and T can't be
cast to Comparable).

The HashSets also assume that you're checking equality via .equals (as
opposed to using ==), and that .hashCode() is correctly implemented,
i.e., consistent with equals.

--
Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth
From: Mike Schilling on 1 Mar 2010 19:36
Tom Anderson wrote:

>
> boolean areAllUnique(List<T> l) {
> l = new ArrayList<T>(l);
> Collections.sort(l);
> T prev = null;
> for (T cur: l) {
> if (cur.equals(prev)) return false;
> prev = cur;
> }
> return true;
> }
>
> Which is pretty different, but still built around the idea of sorting
> and then detecting duplicates by comparing adjacent elements, as sort
> -u does.

This assumes the elements can be ordered, so it fails to be general for the
same reason that the TreeSort solution does.


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