9
Last seen 10 years ago
Member for 10 years, 6 months, 16 days
Difficulty Normal
First-diptiranjan
1
instead of comparing with ord() you could have just checked each letter with letter.isidigit(), letter.isupper(), and letter.islower()
your solution will not work for special letters, like german "umlauts", etc...
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simple and readable solution. however, as you might have noticed, the same schema is used for the roman literals for the hundreds, tens, and ones. putting this scheme into a function would considerably reduce the length of your solution.
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nice and concise solution. i like the idea of joining the FIRST_TEN and SECOND_TEN lists..
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First-teddyxing
simple and good solution; just a short notice: instead of writing
"n=length/2
n=int(n)"
you could just have used integer division: n = length // 2
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First-FeixiangYin
just a hint: use the list.count() function to save you some work ;-)
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wasn't the question to just find one way through the labyrinth, not necessarily the shortest one? using dijkstra for this seems a bit like "shooting with cannons on sparrows" ;-)
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