Azzera filtri
Azzera filtri

What is wrong with the code?

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GEORGIOS BEKAS
GEORGIOS BEKAS il 23 Gen 2018
Commentato: GEORGIOS BEKAS il 24 Gen 2018
I am trying to find the logest subsequence of 1s in a string. I am doing something wrong.
s='0101010111000101110001011100010100001110110100000000110001001000001110001000111010101001101100001111'
c=[]
counter = 0
for i = 2:length(s)
while str2num(s(i)) == 1 && str2num(s(i)) == str2num(s(i-1))
counter = counter+1
c = [c,counter]
if str2num(s(i)) ==0
counter = 0
end
end
end
  2 Commenti
Walter Roberson
Walter Roberson il 23 Gen 2018
hint: instead of doing str2num() and comparing to 1, you can just compare s(i) == '1', and you can compare s(i) == s(i-1)
GEORGIOS BEKAS
GEORGIOS BEKAS il 23 Gen 2018
does not work

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Risposta accettata

Birdman
Birdman il 24 Gen 2018
Use regexp.
regexp(s,'1*','match')
and you will find that the longest subsequence consists of 4 elements.
  1 Commento
GEORGIOS BEKAS
GEORGIOS BEKAS il 24 Gen 2018
if isempty(y) == 1 y = 0 else y=length(y{max(length(y))} ) end

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Più risposte (1)

Walter Roberson
Walter Roberson il 24 Gen 2018
You have
counter = 0;
for i = 2:length(s)
while str2num(s(i)) == 1 && str2num(s(i)) == str2num(s(i-1))
counter = counter+1
c = [c,counter]
if str2num(s(i)) ==0
counter = 0
end
end
Trace it through.
Start with i = 2.
s(2) == 1 but s(1) is not 1, so end the while.
Go on to i = 3. s(3) == 0, so end the while.
Go on to i = 4. s(4) == 1, but s(3) is not 1, so end the while.
Go on to i = 5.... etc. You keep ending the while immediately until...
i = 9. s(9) == 1 and s(9) and s(8) are both 1, so enter the while loop.
Inside the while loop, increment counter to 1 and adjust c. s(9) is still not 0 so do not reset counter to 0. Continue around in the while loop.
i is still 9. s(9) and s(8) are still both 1, so enter the while loop. Inside the while loop, increment counter to 2 and adjust c. s(9) is still not 0, so do not reset counter to 0. Continue in the while loop.
i is still 9. s(9) and s(8) are still both 1, so enter the while loop. Inside the while loop, increment counter to 3 and adjust c. s(9) is still not 0, so do not reset counter to 0. Continue in the while loop.
...
ummm... when do we end the while loop? The while loop tests s(i) and s(i-1) but does not change either location and does not change i, so once entered, the while loop will never end.
  2 Commenti
GEORGIOS BEKAS
GEORGIOS BEKAS il 24 Gen 2018
does not work
Walter Roberson
Walter Roberson il 24 Gen 2018
What did you change your code to?

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