Get all Indexes of the set bits in a BitSet

Question

I'm looking for a fast algorithm with gives me all the indexes of the set bits in a BitSet object. This is slow:

BitSet bitSet = ...
Collection<Integer> indexes = new ArrayList<Integer>(bitSet.cardinality());
int nextSetBit = bitSet.nextSetBit(0);
for (int i = 0; i < bitSet.cardinality(); ++i ) {
    indexes.add(nextSetBit);
    nextSetBit = bitSet.nextSetBit(nextSetBit + 1);
}
...

Any help is appreciated!

Please read this nice article http://java-performance.info/bit-sets/ — Shmil The Cat, Mar 13 '13 at 16:19

score 20 · Answer 1 · answered Mar 13 '13 at 18:05

20

No need to use bitSet.cardinality() at all:

for (int i = bitSet.nextSetBit(0); i != -1; i = bitSet.nextSetBit(i + 1)) {
    indexes.add(i);
}

answered Mar 13 '13 at 18:05

Louis Wasserman

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score 7 · Answer 2 · answered Jan 05 '16 at 20:17

As specified in BitSet#nextSetBit(int) javadocs :

//To iterate over the true bits in a BitSet, use the following loop:
for (int i = bs.nextSetBit(0); i >= 0; i = bs.nextSetBit(i+1)) {
     // operate on index i here
     if (i == Integer.MAX_VALUE) {
         break; // or (i+1) would overflow
     }
 }

score 7 · Answer 3 · edited Jun 20 '20 at 09:12

7

As of Java 8 you can use BitSet's stream() method.

IntStream stream()

Returns a stream of indices for which this BitSet contains a bit in the set state.

edited Jun 20 '20 at 09:12

Community

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answered Nov 05 '18 at 22:09

Luigi Plinge

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score 0 · Answer 4 · answered Aug 17 '22 at 23:22

0

Kotlin solution:

val indexes = MutableList<Int>()
var i = bitSet.nextSetBit(0)
while (i != -1) {
    indexes.add(i)
    i = bitSet.nextSetBit(i + 1)
}

answered Aug 17 '22 at 23:22

naXa stands with Ukraine

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score 0 · Answer 5 · answered Dec 05 '22 at 09:18

0

bitset.toString() will return all the true bit index
like
10010100
{2, 4, 7}
so you can transfer{2, 4, 7}to java list

answered Dec 05 '22 at 09:18

jacky

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score -1 · Answer 6 · answered Mar 13 '13 at 16:24

-1

Change the loop (you increase the complexity to O(N^2) because you call cardinality() in each loop iteration):

for (int e = bitSet.cardinality(), i = 0; i < e; ++i ) {
    indexes.add(nextSetBit);
    nextSetBit = bitSet.nextSetBit(nextSetBit + 1);
}

answered Mar 13 '13 at 16:24

Durandal

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@myborobudur If it didn't help your perceived slowness most likely is *not* caused by the BitSet (either its so small it doesn't matter *or* the autoboxing to Integer overshadows it by far). Use a profiler to measure. – Durandal Mar 13 '13 at 17:02
It is better to use the solution that does not call cardinality() – Phil Jan 22 '17 at 17:09

Get all Indexes of the set bits in a BitSet

6 Answers6