Effacer les filtres
Effacer les filtres

What is wrong with the code?

2 vues (au cours des 30 derniers jours)
GEORGIOS BEKAS
GEORGIOS BEKAS le 23 Jan 2018
Commenté : GEORGIOS BEKAS le 24 Jan 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 commentaires
Walter Roberson
Walter Roberson le 23 Jan 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 le 23 Jan 2018
does not work

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Réponse acceptée

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

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Plus de réponses (1)

Walter Roberson
Walter Roberson le 24 Jan 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 commentaires
GEORGIOS BEKAS
GEORGIOS BEKAS le 24 Jan 2018
does not work
Walter Roberson
Walter Roberson le 24 Jan 2018
What did you change your code to?

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