在Scala中有效地重复一个字符/字符串n次 [英] Efficiently repeat a character/string n times in Scala

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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