[英]Performance of iterating a list by recursively examining the tail
我決定嘗試通過做一些CodinGame挑戰來學習Haskell(所以這個問題是超級初學者級的東西,我敢肯定)。 其中一個需要在整數列表中搜索任意兩個值之間的最小差異。 我以前通過這樣做在Clojure中解決了它:
(ns Solution
(:gen-class))
(defn smallest-difference [values]
(let [v (sort values)]
(loop [[h & t] v curr-min 999999]
(if (nil? t) curr-min
(let [dif (- (first t) h)]
(recur t (if (> curr-min dif) dif curr-min)))))))
(defn -main [& args]
(let [horse-strengths (repeatedly (read) #(read))]
(let [answer (smallest-difference horse-strengths)]
(println answer))))
我嘗試在Haskell中實現相同的解決方案,具體如下:
readHorses :: Int -> [Int] -> IO [Int]
readHorses n h
| n < 1 = return h
| otherwise = do
l <- getLine
let hn = read l :: Int
readHorses (n - 1) (hn:h)
findMinDiff :: [Int] -> Int -> Int
findMinDiff h m
| (length h) < 2 = m
| (h!!1 - h!!0) < m = findMinDiff (tail h) (h!!1 - h!!0)
| otherwise = findMinDiff (tail h) m
main :: IO ()
main = do
hSetBuffering stdout NoBuffering -- DO NOT REMOVE
input_line <- getLine
let n = read input_line :: Int
hPrint stderr n
horses <- readHorses n []
hPrint stderr "Read all horses"
print (findMinDiff (sort horses) 999999999)
return ()
對於Clojure解決方案沒有的大輸入(99999值),這次超時。 然而,它們看起來與我很相似。
至少從表面上看,讀取值並構建列表不是問題,因為在超時之前會打印“Read all horses”。
如何使Haskell版本更高效?
順便說一下,完全替代的實現可以寫成如下。 (偽代碼如下)
拿這份清單
h = [h0, h1, h2 ...
刪除一個元素
drop 1 h = [h1, h2, h3 ...
計算逐點差異
zipWith (-) (drop 1 h) h = [h1-h0, h2-h1, h3-h2, ...
然后采取最低限度。 完整代碼:
minDiff :: [Int] -> Int
minDiff h = minimum (zipWith (-) (drop 1 h) h)
請注意,這將在空列表中崩潰。 另一方面,沒有必要使用9999999
hack。 由於懶惰,它也在恆定的空間中運行。
為了更好的錯誤處理:
minDiff :: [Int] -> Int
minDiff [] = error "minDiff: empty list"
minDiff h = minimum (zipWith (-) (drop 1 h) h)
甚至,更迂腐(但尊重整體):
minDiff :: [Int] -> Maybe Int
minDiff [] = Nothing
minDiff h = Just (minimum (zipWith (-) (drop 1 h) h))
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