Question

Given the byte array:

{255, 3, 5}

which is equivalent to:

{11111111, 00000011, 00000101}

I’d like to get the following result:

{23,22,21,20,19,18,17,16, 9,8, 2,0}

Which is an array of the indices of 1’s in the input array.

What’s the fastest way of doing this in Java?

Update:
I’ve chosen the fastest solution, which @aioobe’s. Here are the test results of a pretty big data test:

@aioobe’s way:

35s 289ms
35s 991ms
36s 174ms

@Martijn’s way:

39s 274ms
39s 879ms
38s 684ms

Thanks you all! I appreciate your help.

Answer 1

What’s the fastest way of doing this in Java?

Presumably by a 256 entry look-up table of the type int[][] in which lut[yourByte] equals the array of indexes for the ones in yourByte.

You then just do something like

for (int i = 0; i < bytes.length; i++)
    for (int indexes : lut[bytes[i]])
        appendToResult(indexes + (bytes.length - 1 - i) * 8);