我正在尝试编写一个将数字转换为等效字符串的程序。 IE 987,654,321 =九十八七百万六五十四数千三百二十一。我从控制台读取值，我想迭代从i = 0到string.length的while循环
示例
var s = "987654321"
var characters = s.toString;
var i = 0;
while(i < characters.length){
do something
}
我的代码在
之下 object project1 {
def main(args: Array[String]) {
println("Enter a number")
val s = Console.readLine
println(eval(s));
}
def tens(c: Char) : String = (
if (c.toInt == 9){
return "Ninety";
} else if (c.toInt == 8){
return "Eighty";
} else if(c.toInt == 7){
return "Seventy";
} else if(c.toInt == 6){
return "Sixty";
} else if(c.toInt == 5){
return "Fifty";
} else if(c.toInt == 4){
return "Fourty";
} else if(c.toInt == 3){
return "Thirty";
} else if(c.toInt == 2){
return "Twenty";
} else{
return "";
}
)
def everyThingElse(c : Char): String = (
if (c.toInt == 9){
return "Nine";
} else if (c.toInt == 8){
return "Eight";
} else if(c.toInt == 7){
return "Seven";
} else if(c.toInt == 6){
return "Six";
} else if(c.toInt == 5){
return "Five";
} else if(c.toInt == 4){
return "Four";
} else if(c.toInt == 3){
return "Three";
} else if(c.toInt == 2){
return "Two";
} else if(c.toInt == 1){
return "One";
} else{
return "";
}
);
def eval(s: String): String = {
val characters = s.toCharArray;
var word = "";
var i = 0;
while( i < characters.length ){
if((i == 14) || (i == 11) || (i == 8) || (i == 5) || (i == 2)){
word = word + everyThingElse(characters(i));
word = word + " Hundred ";
i += 1;
} else if ((i == 13) || (i == 10) || (i == 7) || (i == 4) || (i == 1)){
if(characters(i).toInt != 1){
word = word + tens(characters(i));
i += 1;
} else{
i += 1;
if(characters(i).toInt == 9){
word = word + " Nineteen ";
} else if(characters(i).toInt == 8){
word = word + " Eighteen ";
} else if(characters(i).toInt ==7){
word = word + " Seventeen ";
} else if(characters(i).toInt == 6){
word = word + " Sixteen ";
} else if(characters(i).toInt == 5){
word = word + " Fifteen ";
} else if(characters(i).toInt == 4){
word = word + " Fourteen ";
} else if(characters(i).toInt == 3){
word = word + " Thirdteen ";
} else if(characters(i).toInt == 2){
word = word + " Twelve ";
} else if(characters(i).toInt == 1){
word = word + " Ten ";
}
i += 1;
}
} else if (i == 9){
word = word + everyThingElse(characters(i));
word = word + " Billion ";
i += 1;
} else if(i == 6){
word = word + everyThingElse(characters(i));
word = word + " Million ";
i += 1;
} else if (i == 12){
word = word + everyThingElse(characters(i));
word = word + " Trillion ";
i += 1;
} else if (i == 0){
word = word + everyThingElse(characters(i));
i += 1;
}
}
}
}
答案 0 :(得分：7)
我决定将输出作为选项很有趣，所以我创建了：
object Longhand {
val small = (
" one two three four five six seven eight nine ten eleven twelve " +
"thirteen fourteen fifteen sixteen seventeen eighteen nineteen"
).split(' ')
val mid = "twen thir for fif six seven eigh nine".split(' ').map(_ + "ty")
lazy val big: Stream[String] = {
Stream("", "thousand") #:::
"m b tr quadr quint sext sept oct non dec".split(' ').map(_+"illion").toStream #:::
big.drop(1).map(_ + "-decillion")
}
def duo(n: Int) = {
if (n >= 20) List(mid(n/10 - 2), small(n % 10)).filterNot(_.isEmpty)
else List(small(n)).filterNot(_.isEmpty)
}
def trio(n: Int, and: Boolean = false) = {
val tens = duo(n % 100) match {
case Nil => Nil
case x => if (and) "and" :: x else x
}
List(small(n/100)).filterNot(_.isEmpty).map(_ + " hundred") ::: tens
}
def triples(s: String) = s.reverse.grouped(3).map(_.reverse.toInt).toList
def apply(s: String): String = {
val and = (s.length>2) #:: Stream.continually(false)
val tri = (triples(s) zip and).map{ case (s,a) => trio(s,a) }
val all = (tri zip big).collect{ case (t,b) if !t.isEmpty => t :+ b }
all.reverse.flatten.filterNot(_.isEmpty).mkString(" ")
}
def apply(l: Long): String = if (l<0) "minus "+apply(-l) else apply(l.toString)
def apply(b: BigInt): String = if (b<0) "minus "+apply(-b) else apply(b.toString)
}
处理任意长度的字符串。它拥有一切和厨房水槽（更适合代码高尔夫而不是教学目的），但也许你可以从中得到一些指示。特别是，你应该尝试使用列表（数组，向量，某些东西），以避免一遍又一遍地反复重复自己....
有没有想过最大的无符号长long（Long）写出来了什么？
scala> Longhand(BigInt(2).pow(64)-1)
res32: String = eighteen quintillion four hundred forty six quadrillion
seven hundred forty four trillion seventy three billion seven hundred nine million
five hundred fifty one thousand six hundred and fifteen
编辑：等等，如果你想解决反问题怎么办？只需在apply之后和右括号之前插入：
val bigset = big.take(12).filterNot(_.isEmpty).toSet
val smallset = small.filterNot(_.isEmpty).toSet
val midset = mid.toSet
val valid = (bigset ++ smallset ++ mid) + "hundred"
val bignum = Iterator.iterate(BigInt(1000))(_ * 1000).take(11).toArray
def unapply(text: String): Option[BigInt] = {
val bits = text.toLowerCase.trim.split(" +|-").filter(x => !x.isEmpty && x!="and")
if (!bits.forall(valid contains _)) return None
val parts = bits.map{ case s =>
if (bigset contains s) Right(bignum(big.indexOf(s)-1))
else if (smallset contains s) Left(small.indexOf(s))
else if (midset contains s) Left(20 + 10*mid.indexOf(s))
else Left(100)
}.foldLeft(List[Either[Int,BigInt]]()){ (xs,x) =>
x match {
case Left(i) => xs match {
case Left(j) :: more =>
if (i==100) {
if (j>=10) return None else Left(i*j) :: more
}
else if (i > 10 && (j%100) != 0) return None
else if ((j%10) != 0) return None
else Left(i+j) :: more
case _ => Left(i) :: xs
}
case Right(n) => xs match {
case Right(m) :: more => Right(n*m) :: more
case Left(j) :: more => Right(n*j) :: more
case Nil => Right(n) :: Nil
}
}
}.collect{ case Right(n) => n; case Left(i) => BigInt(i) }
Some(parts.foldLeft(BigInt(0)){ (acc,x) =>
if (acc > x) return None else acc+x
})
}
现在你可以：
scala> "five million eight hundred and one thousand and sixty-two"
res0: String = five million eight hundred and one thousand and sixty-two
scala> res0 match { case Longhand(n) => n }
res1: BigInt = 5801062
scala> Longhand.unapply("one and hundred billion five seven million")
res2: Option[BigInt] = None
（在某些情况下，它会接受实际上没有正确格式化的内容，但拒绝格式错误的数字非常好。）