I need to represent an array of integers using BitSet. Can somebody explain me the logic required to do this ?
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I think the logic would be: Run through the integer array, test every bit and set this bit in the bitset like bitset.set(array_pos+bit_pos)InsertNickHere– InsertNickHere2010-05-26 06:30:39 +00:00Commented May 26, 2010 at 6:30
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3 Answers
You can represent a set of integers using BitSet, but not an arbitrary array. You will lose information about order and repetitions.
Basically, set the nth bit of the BitSet if and only if n appears in your set of integers.
BitSet bitSet = new BitSet();
int[] setOfInts = new int[] { /* Your array here */ };
for (int n : setOfInts) {
bitSet.set(n);
}
Comments
I think the logic would be: Run through the integer array, test every bit and set this bit in the bitset like bitset.set(array_pos+bit_pos)
1 Comment
Noel M
int nums[] = {2, 1}. When arrayPos is 0, you set 2 (0 + 2). When arrayPos is 1, you set 2 again (1 + 1).
first thought:
use BigInteger and create it like: new BigInteger(int value, int base). Then you can toString() it, and then create BitSet using that String(don't know how to do it without analyzing the string, however).
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didn't read it right. That method only helps you to create an array of BitSet, not the whole BitSet that contains the whole array.
I don't know how to make array of integers to one bitSet. I guess you will need some kind of delimeters, but how to make good delimeter in binary - that's a good question.
