您可以对多个元素进行模式匹配,然后将过滤器应用于中间的元素。
eitherside :: (Int->Bool) -> [Int] -> [Int]
eitherside f (i1:i2:i3:is) = if (f i2)
then eitherside f is
else i1 : (eitherside f (i2:i3:is))
eitherside f is = is
*Main> eitherside (==4) [1..10]
[1,2,6,7,8,9,10]
*Main> eitherside (==5) [1..10]
[1,2,3,7,8,9,10]
*Main> eitherside (==6) [1..10]
[1,2,3,4,8,9,10]
不是这样的(我原来的帖子):
eitherside :: (Int->Bool) -> [Int] -> [Int]
eitherside f (i1:i2:i3:is) = if (f i2)
then eitherside f is
else [i1,i2,i3] ++ (eitherside f is)
eitherside f is = is
*Main> eitherside (==5) [1..10]
[1,2,3,7,8,9,10]
这个坏的碰巧对 5 有效,但对 6 失败,因为我在“else”分支中跳过了它。