Re: How to check variables for uniqueness ?

From:
Patricia Shanahan <pats@acm.org>
Newsgroups:
comp.lang.java.programmer
Date:
Thu, 21 Dec 2006 11:30:33 GMT
Message-ID:
<tfuih.2029$X72.297@newsread3.news.pas.earthlink.net>
John Ersatznom wrote:

Andrew Thompson wrote:

Patricia Shanahan wrote:

Andrew Thompson wrote:

krisl...@gmail.com wrote:
...

I have eight variables : var1, var2... var 8. All types String.
How to check that each variables has unique values ?


One way would be to create a Map, iterate the
var's and if not present in the map, add the value
as a key, else return false.


...

Any particular reason for Map, rather than Set?


You mean besides, 'lack of enough consultation
of the relevant docs.'? ;-)

Note that the result of a Set add call is true if, and only if, the
value is not already in the Set.


A Set sounds the go - it is just right for this task.


HashSet<String> foo = new HashSet<String>();
foo.add(var1);
foo.add(var2);
foo.add(var3);
foo.add(var4);
foo.add(var5);
foo.add(var6);
foo.add(var7);
foo.add(var8);
if (foo.size() < 8)
    duplicateExists();
else
    duplicateDoesNotExist();

If you actually need to identify the specific duplicate pairs, you need
to compare them one by one -- 1 with all the others, 2 with all the
higher-numbered ones, and so on up to 7 and 8, using equals().


To save repititious writing, I'm going to assume the strings are in an
array. The equivalent of your code would be:

HashSet<String> foo = new HashSet<String>();
for(String v:vars){
   foo.add(v);
}
if (foo.size() < vars.length)
     duplicateExists();
else
     duplicateDoesNotExist();

You can simplify finding specific duplicates by checking the foo.add
results:

HashSet<String> foo = new HashSet<String>();
for(int i=0; i<vars.length; i++){
   if(!foo.add(vars[i]){
     for(int j=0; j<i; j++){
       if(vars[i].equals(vars[j])){
         reportDuplicate(i,j);
       }
     }
   }
}

A true result from foo.add means the string was actually added to the
set, so it has no duplicate with a lower index.

Patricia

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