鉴于您的示例,我会编写如下所示的函数:
partialAdd :: [Int] -> [Int] -> [Int]
partialAdd ls seps = foldr f [] ls
where
f a [] = [a]
f a (x:xs)
| a `elem` seps = a:x:xs
| otherwise = (x+a):xs
*Main> partialAdd [1,2,3,4,5,6] [2,5]
[3,12,6]
顺便提一句。我认为您问题中的解决方案似乎并不像您在示例中指定的那样有效(或者我误解了某些内容):
partial_add :: [Int] -> [Int] -> Int -> [Int]
partial_add [] _ count = []
partial_add (a:x) list count | elem a list = count:partial_add x list 0
| otherwise = partial_add x list (count+a)
*Main> partial_add [1,2,3,4,5,6] [2,5] 0
[1,7]
但它很容易修复以适用于您的示例:
partial_add :: [Int] -> [Int] -> Int -> [Int]
partial_add [] _ count = [count]
partial_add (a:x) list count | elem a list = (count+a):partial_add x list 0
| otherwise = partial_add x list (count+a)
*Main> partial_add [1,2,3,4,5,6] [2,5] 0
[3,12,6]