Haskell-检查列表中是否有重复的元素
这是一项我基本上已经完成的大学家庭作业，我不是来这里寻求答案的 给定的任务是查找给定的元素是否在列表中出现多次。我尝试的算法是创建将用作计数器的Haskell-检查列表中是否有重复的元素,haskell,Haskell,这是一项我基本上已经完成的大学家庭作业，我不是来这里寻求答案的 给定的任务是查找给定的元素是否在列表中出现多次。我尝试的算法是创建将用作计数器的countdup，并计算在列表中找到该元素的次数。如果countDups大于1，则ismembertweeth（应返回Bool）将为True，否则为False 是的，我是Haskell的新手，所以如果这是一种完全可怕的实现方式，我很抱歉 countDups x [] = 0 countDups x (y:ys) | x == y = 1 +
countdupcountDupsismembertweethBoolTrueFalsecountDups x [] = 0
countDups x (y:ys)
| x == y = 1 + countDups x ys
| otherwise = countDups x ys
isMemberTwice x [] = False --base case; empty list
isMemberTwice x (y: ys)
| countDups > 1 = True
| otherwise False
parse error (possibly incorrect indentation or mismatched brackets)
Failed, modules loaded: none.
isMember _ [] = 0
isMember a (x:xs)
| (a == x) = 1
| otherwise isMember a xs
isMemberTwice _ [] = False
isMemberTwice a (x:xs)
| (a == x) = if ((1 + isMember a (x:xs)) > 1) then True
| otherwise isMemberTwice a xs
ismembers isMember x [] = ???
isMember x (y : ys)
| x == y = ???
| otherwise = ???
isMemberTwice x [] = ???
isMemberTwice x (y : ys)
| x == y = ???
| otherwise = ???
isMemberisMemberisMember两次 isMember x [] = ???
isMember x (y : ys)
| x == y = ???
| otherwise = ???
isMemberTwice x [] = ???
isMemberTwice x (y : ys)
| x == y = ???
| otherwise = ???
ismembers isMember x [] = ???
isMember x (y : ys)
| x == y = ???
| otherwise = ???
isMemberTwice x [] = ???
isMemberTwice x (y : ys)
| x == y = ???
| otherwise = ???
isMemberisMemberisMember两次 isMember x [] = ???
isMember x (y : ys)
| x == y = ???
| otherwise = ???
isMemberTwice x [] = ???
isMemberTwice x (y : ys)
| x == y = ???
| otherwise = ???
在最后一个
False=ifelseisMember:：…OrdifelseisMember:：…OrdisMemberisMember