String manipulations: transform "a-b-c" into "a(b(c))"

Question

function dashesToParentheses(str) {
 var list = str.split('-');
 return str.replace(/-/g, '(') + repeatString(')', list.length - 1);
}
function repeatString(str, times) {
 if (times == 1)
 return str;
 return new Array(times + 1).join(str);
}
dashesToParentheses('a-b-c') // "a(b(c))"
dashesToParentheses('a-b') // "a(b)"
dashesToParentheses('a') // "a"
dashesToParentheses('') // ""

dashesToParentheses works correct. Can I make it simpler or/and faster?

Answer 1

Having split the string you can join it with brackets instead of replacing them. You could optionally choose to remove the repeatString function and your +/- 1, but it does make a lot of sense the way you have it.

function dashesToParentheses(str) {
 var list = str.split('-');
 return list.join('(') + Array(list.length).join(')');
}

Answer 2

Well, you could remove the RegExp. But whether that helps performance is anyone's guess.

var d2p = function(s){
 var one=[], two=[], a=s.split('-');
 one.push(a.shift());
 a.forEach(function(part){
 one.push('(' + part);
 two.push(')');
 });
 return one.join('') + two.join('');
};

If you don't mind having the result fully parenthesised, then you can do this:

var d2p = function(s){
 return s.split('-').reduceRight(function(whole, part){
 return '(' + part + whole + ')';
 }, '');
};

Answer 3

The first term is obvious: Just replace all dashes by opening parentheses. The second replaces all dashes by closing parentheses while dropping everything else.

function dashesToParentheses(str) {
 return str.replace(/-/g, "(") + str.replace(/[^-]*-[^-]*/g, ")");
}