scala Scala反向字符串
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/7700399/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Scala reverse string
提问by Dzhu
I'm a newbie to scala, I'm just writing a simple function to reverse a given string:
我是 Scala 的新手，我只是在编写一个简单的函数来反转给定的字符串：
def reverse(s: String) : String
for(i <- s.length - 1 to 0) yield s(i)
the yield gives back a scala.collection.immutable.IndexedSeq[Char], and can not convert it to a String. (or is it something else?)
yield 返回一个 scala.collection.immutable.IndexedSeq[Char]，并且不能将其转换为 String。（或者是别的什么？）
how do i write this function ?
我怎么写这个函数？
回答by om-nom-nom
Note that there is already defined function:
请注意，已经定义了函数：
scala> val x = "scala is awesome"
x: java.lang.String = scala is awesome
scala> x.reverse
res1: String = emosewa si alacs
But if you want to do that by yourself:
但如果你想自己做：
def reverse(s: String) : String =
(for(i <- s.length - 1 to 0 by -1) yield s(i)).mkString
or (sometimes it is better to use until, but probably not in that case)
或（有时最好使用until，但在这种情况下可能不是）
def reverse(s: String) : String =
(for(i <- s.length until 0 by -1) yield s(i-1)).mkString
Also, note that if you use reversed counting (from bigger one to less one value) you should specify negative step or you will get an empty set:
另外，请注意，如果您使用反向计数（从大一到小一值），您应该指定负步长，否则您将得到一个空集：
scala> for(i <- x.length until 0) yield i
res2: scala.collection.immutable.IndexedSeq[Int] = Vector()
scala> for(i <- x.length until 0 by -1) yield i
res3: scala.collection.immutable.IndexedSeq[Int] = Vector(16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
回答by Luigi Plinge
Here's a short version
这是一个简短的版本
def reverse(s: String) = ("" /: s)((a, x) => x + a)
edit: or even shorter, we have the fantastically cryptic
编辑：或者更短，我们有非常神秘的
def reverse(s: String) = ("" /: s)(_.+:(_))
but I wouldn't really recommend this...
但我真的不会推荐这个......
回答by huynhjl
As indicated by om-nom-nom, pay attention to the by -1(otherwise you are not really iterating and your result will be empty). The other trick you can use is collection.breakOut.
正如 om-nom-nom 所指示的那样，注意by -1（否则你不是真正的迭代，你的结果将是空的）。您可以使用的另一个技巧是 collection.breakOut.
It can also be provided to the forcomprehension like this:
它也可以for像这样提供给理解：
def reverse(s: String): String =
(for(i <- s.length - 1 to 0 by -1) yield s(i))(collection.breakOut)
reverse("foo")
// String = oof
The benefit of using breakOutis that it will avoid creating a intermediate structure as in the mkStringsolution.
使用的好处breakOut是它将避免在mkString解决方案中创建中间结构。
note: breakOutis leveraging CanBuildFromand builders which are part of the foundation of the redesigned collection library introduced in scala 2.8.0
注意：breakOut正在利用CanBuildFrom和构建器，它们是 Scala 2.8.0 中引入的重新设计的集合库的基础的一部分
回答by Thomas Lockney
You could also write this using a recursive approach (throwing this one in just for fun)
您也可以使用递归方法编写它（只是为了好玩而将其放入）
def reverse(s: String): String = {
if (s.isEmpty) ""
else reverse(s.tail) + s.head
}
回答by Venkat Sudheer Reddy Aedama
All the above answers are correct and here's my take:
以上所有答案都是正确的，这是我的看法：
scala> val reverseString = (str: String) => str.foldLeft("")((accumulator, nextChar) => nextChar + accumulator)
reverseString: String => java.lang.String = <function1>
scala> reverseString.apply("qwerty")
res0: java.lang.String = ytrewq
回答by Igor Dorokhov
def rev(s: String): String = {
val str = s.toList
def f(s: List[Char], acc: List[Char]): List[Char] = s match {
case Nil => acc
case x :: xs => f(xs, x :: acc)
}
f(str, Nil).mkString
}
