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adding-ordinal-indicator-suffixes-to-numbers.md

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Description

Finish the function numberToOrdinal, which should take a number and return it as a string with the correct ordinal indicator suffix (in English). That is:

  • numberToOrdinal(1) ==> '1st'
  • numberToOrdinal(2) ==> '2nd'
  • numberToOrdinal(3) ==> '3rd'
  • numberToOrdinal(4) ==> '4th'
  • ... and so on

For the purposes of this kata, you may assume that the function will always be passed a non-negative integer. If the function is given 0 as an argument, it should return '0' (as a string).

To help you get started, here is an excerpt from Wikipedia's page on Ordinal Indicators:

  • st is used with numbers ending in 1 (e.g. 1st, pronounced first)
  • nd is used with numbers ending in 2 (e.g. 92nd, pronounced ninety-second)
  • rd is used with numbers ending in 3 (e.g. 33rd, pronounced thirty-third)
  • As an exception to the above rules, all the "teen" numbers ending with 11, 12 or 13 use -th (e.g. 11th, pronounced eleventh, 112th, pronounced one hundred [and] twelfth)
  • th is used for all other numbers (e.g. 9th, pronounced ninth).

Kata's link: Adding ordinal indicator suffixes to numbers

Best Practices

First:

function numberToOrdinal(n) {
  var suffix = "th";
  if (n == 0) suffix = "";
  if (n % 10 == 1 && n % 100 != 11) suffix = "st";
  if (n % 10 == 2 && n % 100 != 12) suffix = "nd";
  if (n % 10 == 3 && n % 100 != 13) suffix = "rd";
  return n + suffix;
}

Second:

function numberToOrdinal(n) {
  s = n.toString();
  if ( s.match(/\d*(11|12|13)$/) ) return s + 'th';
  if ( s.match(/\d*(1)$/) ) return s + 'st';
  if ( s.match(/\d*(2)$/) ) return s + 'nd';
  if ( s.match(/\d*(3)$/) ) return s + 'rd';
  if ( s === '0' ) return s;
  return s + 'th';
}

My solutions

function numberToOrdinal(n) {
  // Finish me
  if(n === 0) {
    return '0';
  }

  if (n > 0 && Number.isSafeInteger(n)) {

      if(n >= 10 && n <= 20) {
        return '' + n +'th';
      }

      if(/^\d+(1\d+)$/.test(n)) {
          return  '' + n + 'th';
       }

      if(/^\d?1$/.test(n) || /^\d+(0+1)$/.test(n)) {
        return '' + n + 'st';
      }
      if(/^\d?2$/.test(n)) {
        return  ''+ n + 'nd';
      }
      if(/^\d?3$/.test(n) || /^\d+(\d+3)$/.test(n)) {
        return  ''+ n + 'rd';
      }
      
      if(/^\d{1,2}/.test(n)){
         return ''+ n + 'th';
      }
   }
}