Count of consecutive duplicate integers in a list with frequency

maddie :

I want to count the number of consecutive repeating numbers from a list of bytes and show them as two integers arrays:

• The first array contains the non-repeating integer values.
• The second array contains the consequent repetition counts.

So, for an input like this:

Byte[] bytes = new Byte[] {2, 2, 2, 0, 0, 0, 0, 2, 2, 0, 0, 2};

I expect an output like this:

integers  -[2, 0, 2, 0, 2]
frequency -[3, 4, 2, 2, 1]

Which is basically a compressed view of the input. The output says integer 2 repeats 3 times followed by 0 that repeats 4 times followed by 2 that repeats 2 times and so on..

I have written the below code.

 List<Byte> integers = new ArrayList<>();
 List<Integer> frequencies = new ArrayList<>();

 for (int i=0; i < bytes.size() - 1; i++) {
     Byte current = bytes.get(i);
     Byte next = bytes.get(i+1);
     if (current == next) {
         count ++;
         // if all bytes are of the same type
         if (count == bytes.size() || i == bytes.size() - 2) {
             integers.add(current);
             frequencies.add(count);
         }
         continue;
         integers.add(current);
         frequencies.add(count);
         count = 1;
     }
 }
 System.out.println("integers " +  integers + " - frequency " + frequencies);

This code works for most cases. But I am missing some edge cases. Like for the example input, the output is missing to reach to the last element 2. The output from my code for input is -

integers  -[2, 0, 2, 0]
frequency -[3, 4, 2, 2]

I am adding a bunch of if statements to cover all corner cases but I want to know if there is a cleaner solution to this?

Can Bayar :

I didn't try running this code on an IDE but I think this should suffice:

int count = 1;
int index = 0;

byte current = bytes[index];
while (index < bytes.length - 1) {
    index++;
    if (bytes[index] == current) {
        count++;
    } else {
        integers.add(current);
        frequencies.add(count);
        count = 1;
        current = bytes[index];
    }
}

integers.add(current);
frequencies.add(count);