What is wrong with the code?

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GEORGIOS BEKAS
GEORGIOS BEKAS 2018-1-23
评论： GEORGIOS BEKAS 2018-1-24
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 个评论
Walter Roberson
Walter Roberson 2018-1-23
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)

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采纳的回答

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

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更多回答（1 个）

Walter Roberson
Walter Roberson 2018-1-24
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.

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