I will explain first what I mean by "complementing integer value excluding the leading zero binary bits" (from now on, I will call it Non Leading Zero Bits complement or NLZ-Complement for brevity).
For example, there is integer number 92. the binary number is 1011100. If we perform normal bitwise-NOT or Complement, the result is: -93 (signed integer) or 11111111111111111111111110100011 (binary). That's because the leading zero bits are being complemented too.
So, for NLZ-Complement, the leading zero bits are not complemented, then the result of NLZ-complementing of 92 or 1011100 is: 35 or 100011 (binary). The operation is performed by XORing the input value with sequence of 1 bits as much as the non-leading zero value. The illustration:
92: 1011100
1111111 (xor)
--------
0100011 => 35
I had made the java algorithm like this:
public static int nonLeadingZeroComplement(int n) {
if (n == 0) {
return ~n;
}
if (n == 1) {
return 0;
}
//This line is to find how much the non-leading zero (NLZ) bits count.
//This operation is same like: ceil(log2(n))
int binaryBitsCount = Integer.SIZE - Integer.numberOfLeadingZeros(n - 1);
//We use the NLZ bits count to generate sequence of 1 bits as much as the NLZ bits count as complementer
//by using shift left trick that equivalent to: 2 raised to power of binaryBitsCount.
//1L is one value with Long literal that used here because there is possibility binaryBitsCount is 32
//(if the input is -1 for example), thus it will produce 2^32 result whom value can't be contained in
//java signed int type.
int oneBitsSequence = (int)((1L << binaryBitsCount) - 1);
//XORing the input value with the sequence of 1 bits
return n ^ oneBitsSequence;
}
I need an advice how to optimize above algorithm, especially the line for generating sequence of 1 bits complementer (oneBitsSequence), or if anyone can suggest better algorithm?
UPDATE: I also would like to know the known term of this non-leading zero complement?
You can get the highest one bit through the Integer.highestOneBit(i)
method, shift this one step left, and then subtract 1. This gets you the correct length of 1
s:
private static int nonLeadingZeroComplement(int i) {
int ones = (Integer.highestOneBit(i) << 1) - 1;
return i ^ ones;
}
For example,
System.out.println(nonLeadingZeroComplement(92));
prints
35