jerryu
commented
3 years ago I am also curious, so I added some debugging code to output
- the partial string yet to check at the beginning of the iteration.
- the values of i and count, at beginning and the end of each iteration
The i=i+1 proposed by Naolador seems to have the desirable/expected flow.
Allowing overlapped counting might require walking the pointer back, depending on the flow. To illustrate, I made up a word, 'booookmaker'. Both solutions don't allow overlapping, whereas the second 'o' forms one double with 'o' before and another double with 'o' after.
reset booookmaker 0 0 1 0 reset booookmaker 0 0 1 0 double ooookmaker 1 0 3 1 double ooookmaker 1 0 3 1 double ookmaker 3 1 5 2 double ookmaker 3 1 5 2 reset kmaker 5 2 2 0 | reset kmaker 5 2 6 0 double oookmaker 2 0 4 1 < reset okmaker 4 1 3 0 < double ookmaker 3 0 5 1 < reset kmaker 5 1 4 0 < reset okmaker 4 0 5 0 < reset kmaker 5 0 6 0 < reset maker 6 0 7 0 reset maker 6 0 7 0 reset aker 7 0 8 0 reset aker 7 0 8 0 reset ker 8 0 9 0 reset ker 8 0 9 0 reset er 9 0 10 0 reset er 9 0 10 0
I just don't understand why we use "i = i + 1 - 2*count" to reset the pointer in line 31 if two nearby characters doesn't match. I replaced the line with "i = i + 1" and got the same result.