（Scala）生成一个将数字加1的序列

Question

how do I generate a sequence of size n that would look like this:

n = 3
000
001
002
003
...
010
011
012
...
999

I could use fixed number of for loops for a fixed number of digits but how do i do this with n number of digits in a sequence?

Thank you! ^^

Answer 1

This will give the output you're looking for but as a Seq of String . You can replace 3 with n to get more or less leading zeros.

Range(0,1000).map(n => "%03d".format(n))

scala.collection.immutable.IndexedSeq[String] = Vector(000, 001, 002, 003, 004, 005, 006, 007, 008, 009, 010, 011, ...)

Answer 2

Assuming that:

• n specifies the number of digits in the number
• The range is from 0 to Math.pow(10, n)

A dynamic solution would be:

val n = 3 //Or any number

def upperBound : Int = Math.pow(10, n).toInt
def formatNum(num : Int) : String = s"%0${n}d".format(num)
def range: Range = Range(0, upperBound)

range.map(formatNum)

The main point here being that you can use string interpolation to have variable padding on the numbers using s"..." , where variables are denoted by $var_name or ${var_name}