Popis problému
Celý popis zadania sa nacháza – Hackerrank.
Riešenie
Takýto regulárny výraz overí platnosť prihlasovacieho mena:
^[_\.]\d+[a-zA-Z]*_?$
Vytvoril som riešenie v týchto programovacích jazykoch:
Všetky riešenia sú dostupné aj na mojom GitHub profile.
Java
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import java.util.Scanner; public class AlienUsername { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int testCases = scanner.nextInt(); for (int i = 0; i < testCases; i++) { String input = scanner.next(); if(input.matches("[_\\.]\\d+[a-zA-Z]*_?")) { System.out.println("VALID"); } else { System.out.println("INVALID"); } } scanner.close(); } } |
JavaScript
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function processData(input) { var lines = input.split("\n"); for(var i = 1; i < lines.length; i++) { if(lines[i].match(/^[_\.]\d+[a-z]*_?$/ig)) { console.log("VALID"); } else { console.log("INVALID"); } } } process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { processData(_input); }); |
Scala
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import scala.io.Source object AlienUsername extends App { val usernames = Source.stdin.getLines().drop(1) println(usernames.map(validateUsername).mkString("\n")) def validateUsername(name: String): String = { if (name.matches("[_\\.]\\d+[a-zA-Z]*_?")) return "VALID" else return "INVALID" } } |
Ruby
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n = gets.to_i n.times do name = gets.to_s.strip if /[_\.]\d+[a-zA-Z]*_?/ =~ name puts "VALID" else puts "INVALID" end end |