/**
* Removes from the receiver all elements that are contained in the specified list. Tests for
* identity.
*
* @param other the other list.
* @return <code>true</code> if the receiver changed as a result of the call.
*/
public boolean removeAll(AbstractByteList other) {
// overridden for performance only.
if (!(other instanceof ByteArrayList)) return super.removeAll(other);
/*
* There are two possibilities to do the thing a) use other.indexOf(...)
* b) sort other, then use other.binarySearch(...)
*
* Let's try to figure out which one is faster. Let M=size,
* N=other.size, then a) takes O(M*N) steps b) takes O(N*logN + M*logN)
* steps (sorting is O(N*logN) and binarySearch is O(logN))
*
* Hence, if N*logN + M*logN < M*N, we use b) otherwise we use a).
*/
if (other.size() == 0) {
return false;
} // nothing to do
int limit = other.size() - 1;
int j = 0;
byte[] theElements = elements;
int mySize = size();
double N = (double) other.size();
double M = (double) mySize;
if ((N + M) * cern.jet.math.Arithmetic.log2(N) < M * N) {
// it is faster to sort other before searching in it
ByteArrayList sortedList = (ByteArrayList) other.clone();
sortedList.quickSort();
for (int i = 0; i < mySize; i++) {
if (sortedList.binarySearchFromTo(theElements[i], 0, limit) < 0)
theElements[j++] = theElements[i];
}
} else {
// it is faster to search in other without sorting
for (int i = 0; i < mySize; i++) {
if (other.indexOfFromTo(theElements[i], 0, limit) < 0) theElements[j++] = theElements[i];
}
}
boolean modified = (j != mySize);
setSize(j);
return modified;
}
