Efficiently repeat a character/string n times in S

问题:

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

回答1:

For strings you can just write "abc" * 3, which works via StringOps and uses a StringBuffer behind the scenes.

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.

回答2:

I know this is an old question, another solution would be what I have below

def repeatChar(char:Char, n: Int) : String = {
    var result = ""
    for(_ <- 1 to n){
      result += "" + char
    }
    result

}

result from REPL repeatChar('a', 2) => res1: String = aa