Take a look at java.util.BitSet

Patricia

On Sun, 9 Mar 2008 21:01:55 -0700 (PDT),
"nooneinparticular314159@yahoo.com"
<nooneinparticular314159@yahoo.com> wrote, quoted or indirectly quoted
someone who said :

see http://mindprod.com/jgloss/bitset.html
--

Roedy Green Canadian Mind Products
The Java Glossary
http://mindprod.com

If there are <= 32 bits it fits into an int, and I would do like this:

public static int B1 = 1;
public static int B2 = 2;
public static int B3 = 4;
public static int B4 = 8;

public int bits;

bits = B1; // Set bit 1, all others 0
bits = B1 + B2; // Set bit 1 and 4, all others 0
bits = bits | B3 // Set bit 3, the rest of the bits are unchanged
if ((bits & B1) > 0 ) // Is bit 1 set?
if ((bits & (B1 + B4)) > 0 ) // Is bit 1 OR bit 4 set?

Please note that the syntax is not verified, it only shows the
principle.
I think that, since it's mostly low level operations without class
references, it should performe well.

/Ulf

BitSet looks like just the thing.  HashMap seems overkill.  How are you
looking for the bits? If it's always an index, then arrays (and
presumably BitSet) automatically provide a "perfect hash" for you.

You could roll your own, but I think BitSet would do the same thing as this:

class MyBitSet {
  private ArrayList<Integer> bits = new ArrayList<Integer>();

  public boolean getBit( int i ) {
    int intBits = bits.get( i / 32 );
    return !(intBits & (1<<(i%32)) = 0)
  }
}
// Note: seriously not syntax checked.

And similarly for set... but BitSet might also provide a custom
implementation that is more efficient than ArrayList and Integer (raw
ints, for example.)

(Note: I just peeked at the implementation of BitSet.  I does indeed
appear to use plain arrays of longs.  This should be much more efficient
than trying to use a HashMap.)