Question

我想要一个函数，用字母表中的每个其他字母替换字符串中的字母。示例：

f :: String -> [String]
f = ...

ghci> f "foo"
["aoo","boo,"coo","doo","eoo","goo","hoo"..."fao",fbo","fco","fdo",...."foa","fob","foc","fod"...]

我目前得到了：

alphabet :: [Char]
alphabet = ['a'..'z']

replaced1 :: String -> [String]
replaced1 [] = []
replaced1 (x:xs) = (alphabet++xs) : map (x:) (replaced1 xs)

当然这不起作用，因为它增加了整个字母，但我不知道如何从字母表中只拿一个字符。我希望你能帮助我。

Answer 1

假设您的函数的参数与模式(x:xs)匹配;为了简洁起见，我不会讨论基本案例。

使用列表推导生成一个字符串列表，其中x被字母表中的所有其他字母替换：
```
[ c:xs | c <- alphabet, c /= x ]
```
生成器从字母表中抽取字符c，但谓词禁止c与x相同。
递归地将函数应用于String参数xs的尾部：
```
f xs
```
将字符x添加到2中获得的字符串列表中的每个字符串。
```
map (x:) (f xs)
```

将结果3.附加到1。：

[ c:xs | c <- alphabet, c /= x ] ++ map (x:) (f xs)

全部放在一起：

alphabet :: String
alphabet = ['a'..'z']

f :: String -> [String]
f     [] = []
f (x:xs) = [ c:xs | c <- alphabet, c /= x ] ++ map (x:) (f xs)

测试：

ghci> f "foo"  
["aoo","boo","coo","doo","eoo","goo","hoo","ioo","joo","koo","loo","moo","noo",
"ooo","poo","qoo","roo","soo","too","uoo","voo","woo","xoo","yoo","zoo","fao",
"fbo","fco","fdo","feo","ffo","fgo","fho","fio","fjo","fko","flo","fmo","fno",
"fpo","fqo","fro","fso","fto","fuo","fvo","fwo","fxo","fyo","fzo","foa","fob",
"foc","fod","foe","fof","fog","foh","foi","foj","fok","fol","fom","fon","fop",
"foq","for","fos","fot","fou","fov","fow","fox","foy","foz"]

通过用所有其他字母字符替换字符串中的每个字符获得的字符串列表

1 个答案: