Java获取BitSet交集基数的最快方法

Question

The function below takes two BitSets , makes a copy of the first (it must not be overridden), intersects the copy with the second (bitwise AND) and returns the cardinality of the result.

public int getIntersectionSize(BitSet bits1, BitSet bits2) {
    BitSet copy = (BitSet) bits1.clone();
    copy.and(bits2);
    return copy.cardinality();
}

I'm interested if this code can be sped up? This function is called billion of times so even a microsecond speed up makes sense plus I'm curious about the fastest possible code.

Answer 1

If you're going to use each BitSet several times, it could be worthwhile to create a long array corresponding to each BitSet . For each BitSet :

long[] longs = bitset.toLongArray();

Then you can use the following method, which avoids the overhead of creating a cloned BitSet . (This assumes that both arrays are the same length).

int getIntersectionSize(long[] bits1, long[] bits2) {
    int nBits = 0;
    for (int i=0; i<bits1.length; i++)
        nBits += Long.bitCount(bits1[i] & bits2[i]);
    return nBits;
}

Answer 2

Here is an alternative version, but I'm not sure if it is really faster, depends on nextSetBit .

public int getIntersectionsSize(BitSet bits1, BitSet bits2) {
   int count = 0;
   int i = bits1.nextSetBit(0);
   int j = bits2.nextSetBit(0);
   while (i >= 0 && j >= 0) {
      if (i < j) {
         i = bits1.nextSetBit(i + 1);
      } else if (i > j) {
         j = bits2.nextSetBit(j + 1);
      } else {
         count++;
         i = bits1.nextSetBit(i + 1);
         j = bits2.nextSetBit(j + 1);
      }
   }
   return count;
}

The above is the readable version, hopefully good enough for the compiler, but you could optimize it manually I guess:

public int getIntersectionsSize(BitSet bits1, BitSet bits2) {
   int count = 0;
   for (int i = bits1.nextSetBit(0), j = bits2.nextSetBit(0); i >= 0 && j >= 0; ) {
      while (i < j) {
         i = bits1.nextSetBit(i + 1);
         if (i < 0)
            return count;
      }
      if (i == j) {
         count++;
         i = bits1.nextSetBit(i + 1);
      }
      while (j < i) {
         j = bits2.nextSetBit(j + 1);
         if (j < 0)
            return count;
      }
      if (i == j) {
         count++;
         j = bits2.nextSetBit(j + 1);
      }
   }
   return count;
}

Answer 3

I've been looking for a solution to this recently and here's what I came up with:

int intersectionCardinality(final BitSet lhs, final BitSet rhs) {
    int lhsNext;
    int retVal = 0;
    int rhsNext = 0;

    while ((lhsNext = lhs.nextSetBit(rhsNext)) != -1 &&
            (rhsNext = rhs.nextSetBit(lhsNext)) != -1) {
        if (rhsNext == lhsNext) {
            retVal++;
            rhsNext++;
        }
    }

    return retVal;
}

Perhaps someone would like to take the time to compare the different solutions here and post the results...