Haskell Quiz/Phone Number Words/Solution Jethr0: Difference between revisions
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-- if no result is found, the original string is returned as possibility | -- if no result is found, the original string is returned as possibility | ||
allPoss xs = if null result then [xs] else result | allPoss xs = if null result then [xs] else result | ||
where result = filter (`elem` dictionary) . | where result = filter (`elem` dictionary) . mapM poss $ xs | ||
poss char = encodings!! | poss char = encodings!! digitToInt char | ||
encodings = ["0", "1", "ABC", "DEF", "GHI", "JKL", "MNO", "PQRS", "TUV", "WXYZ"] | encodings = ["0", "1", "ABC", "DEF", "GHI", "JKL", "MNO", "PQRS", "TUV", "WXYZ"] | ||
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-- > phoneNames "1.800.5646.843.4275355.7849" | -- > phoneNames "1.800.5646.843.4275355.7849" | ||
-- ["1-800-JOIN-THE-HASKELL-QUIZ"] | -- ["1-800-JOIN-THE-HASKELL-QUIZ"] | ||
phoneNames = map (concat . intersperse "-") . | phoneNames = map (concat . intersperse "-") . mapM allPoss . splitThem | ||
</haskell> | </haskell> | ||
Latest revision as of 05:17, 13 December 2009
dictionary = map (map toUpper) ["haskell", "join", "the", "quiz"]
-- fetch all possible words represented by a string of consecutive digits
-- the results are checked against the dictionary
-- if no result is found, the original string is returned as possibility
allPoss xs = if null result then [xs] else result
where result = filter (`elem` dictionary) . mapM poss $ xs
poss char = encodings!! digitToInt char
encodings = ["0", "1", "ABC", "DEF", "GHI", "JKL", "MNO", "PQRS", "TUV", "WXYZ"]
splitThem [] = []
splitThem xs = y : rest
where (y,ys) = break (not . isDigit) xs
rest = splitThem . dropWhile (not . isDigit) $ ys
-- > phoneNames "1.800.5646.843.4275355.7849"
-- ["1-800-JOIN-THE-HASKELL-QUIZ"]
phoneNames = map (concat . intersperse "-") . mapM allPoss . splitThem