Question

我正在尝试创建一个函数，该函数将查找给定列表的3个相同且相邻的数字，对于我正在尝试实现的求解器。然后，如果有3个相同且相邻的数字，则将第1个和第3个相同的数字标记为“0”，并将中间值设置为负数。

我想知道为什么这会给我一个错误。：

change xs = chnge xs []
    where
    chnge xs acc
        | length xs <= 2 = [acc]
        | (head xs == xs !! 1) && (head xs == xs !! 2) = [0, (xs !! 1)*(-1), 0] ++ tail xs
        | otherwise = chnge (tail xs) (acc ++ head xs)

Answer 1

由于acc是一个列表，我们不希望在[acc]的第一个后卫中返回chnge，而只是acc;同样在otherwise行，您不希望acc ++ head xs这意味着xs是一个列表列表 - 其第一个成员可以如何附加？相反acc ++ [head xs]所以可能：

change xs = chnge xs [] where
  chnge xs acc
        | length xs <= 2 = acc
        | (head xs == xs !! 1) && (head xs == xs !! 2) = [0, (xs !! 1)*(-1), 0] ++ tail xs
        | otherwise = chnge (tail xs) (acc ++ [head xs])

这看起来有些偏差，但真正的问题是“模式匹配”的缺乏以及head，tail和!!的危险使用。尝试更像这样的东西，也许吧？（虽然它不使用累加器）：

change []     = []
change [x]    = [x]
change [x,y]  = [x,y]
change (x:y:z:ws) | x == y && y == z = 0 : (-y) : 0 : change ws
change (x:xs) =  x : change xs

--  *Main> change [12,12,66,66,66,44,44,99,99,99,76,1]
--  [12,12,0,-66,0,44,44,0,-99,0,76,1]

连续三个案例可以被视为一种模式，因此我们在它们相等的时候提出一个特例。

使用累加器的Haskell递归

1 个答案: