2016-06-30 66 views
1

這是我實現:Haskell的歸併執行編譯,但不返回任何

mergesort :: (Ord a) => [a] -> [a] 
mergesort list = merge (mergesort (left list)) (mergesort (right list)) 
    where 
    left xs = take (div (length xs) 2) xs 
    right xs = drop (div (length xs) 2) xs 
    merge [] ys = ys 
    merge xs [] = xs 
    merge (x:xs) (y:ys) 
     | x <= y = x : merge xs (y:ys) 
     | otherwise = y : merge (x:xs) ys 

代碼編譯,但是當我運行它,我的機器崩潰。我究竟做錯了什麼?

+0

您的*原*代碼是否包含空白的案例? *如果不是*,請回滾該編輯。讓亞歷克知道,以便他可以在他的回答中刪除對該案件的引用。 – Bakuriu

+0

應標記正確的答案。抱歉! – dopatraman

回答

3

你缺少基礎案例 - 所以你得到無限遞歸。嘗試使用[][1]這樣的列表來查看您的示例,您將直接進入該問題。

mergesort :: (Ord a) => [a] -> [a] 
mergesort [] = [] -- < ADDED 
mergesort [x] = [x] -- < ADDED 
mergesort list = merge (mergesort (left list)) (mergesort (right list)) 
    where 
    left xs = take (div (length xs) 2) xs 
    right xs = drop (div (length xs) 2) xs 
    merge [] ys = ys 
    merge xs [] = xs 
    merge (x:xs) (y:ys) 
     | x <= y = x : merge xs (y:ys) 
     | otherwise = y : merge (x:xs) ys 
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