问题描述
我想更有效地执行下列操作:
I would like to do the following more efficiently:
def repeatChar(char:Char, n: Int) = List.fill(n)(char).mkString
def repeatString(char:String, n: Int) = List.fill(n)(char).mkString
repeatChar('a',3) // res0: String = aaa
repeatString("abc",3) // res0: String = abcabcabc
$ b b
推荐答案
对于字符串,你可以写abc* 3
http://www.scala-lang.org/api/current/index.html#scala.collection.immutable.StringOps> StringOps
并使用 StringBuffer
For strings you can just write "abc" * 3
, which works via StringOps
and uses a StringBuffer
behind the scenes.
对于字符我认为你的解决方案是相当合理的,虽然 char.toString * n
可以说更清楚。你有什么理由怀疑 List.fill
版本不能满足你的需要吗?您可以编写自己的方法,使用 StringBuffer
(类似于 *
在 StringOps
),但我建议首先明确目标,然后担心效率只有当你有具体的证据,这是一个问题在你的程序。
For characters I think your solution is pretty reasonable, although char.toString * n
is arguably clearer. Do you have any reason to suspect the List.fill
version isn't efficient enough for your needs? You could write your own method that would use a StringBuffer
(similar to *
on StringOps
), but I would suggest aiming for clarity first and then worrying about efficiency only when you have concrete evidence that that's an issue in your program.
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