Run-length decoder

Question

This replaces numbers with the character that appears next to them ('number' times). Is there a a more elegant or shorter way to do this?

var decode = function(str, result) {
 var regex = /\d+/,
 number = regex.exec(str);
 if (number === null) {
 return str;
 }
 var start = number.index,
 end = number[0].length + start - 1,
 str = str.replace(/\d+/, ""),
 repeat = str.charAt(start);
 result += str.substring(0, start);
 for (var i = 0; i < number[0] - 1; i++) {
 result += repeat;
 }
 result += str.substring(start, str.length);
 number = regex.exec(result);
 if (number === null) {
 return result;
 } else {
 return decode(result, "");
 }
};

---

var str = "bob2b11a";
console.log(decode(str, ""));
// "bobbbaaaaaaaaaaa"

Answer 1

The basic concept

• Utilize string replace with regular expression
• Utilize new Array Creation with join

The code

function decode (str) {
 return str.replace(/(\d+)(\w)/g, 
 function(m,n,c){
 return new Array( parseInt(n,10)+1 ).join(c);
 }
 );
}

• beatgammit: Clever. I forgot about [].join...
• mtwallet: @epascarello please can you tell me in the function, where/how do m, n, c get populated with the values? How does that work? Many thanks in advance.
• epascarello: @mtwallet read documentation for replace developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…
• mtwallet: @epascarello thanks, I looked at w3schools.com/jsref/jsref_replace.asp which didn't have the same level of detail
• epascarello: Avoid w3schools since their documentation lacks in a bunch of areas.

Answer 2

From my answer on Stack Overflow: https://stackoverflow.com/a/13481139/538551

Simply use String.replace():

function decode(str) {
 return str.replace(/(\d+)([a-zA-A])/g, function (match, num, letter) {
 var ret = '', i;
 for (i = 0; i < parseInt(num, 10); i++) {
 ret += letter;
 }
 return ret;
 });
}

• beatgammit: Yeah, but this is the better place, so I wanted to provide it here just for completeness. @epascarello's answer will probably be better.