How do you reverse a string in place in JavaScript?

Question

How do you reverse a string in place (or in-place) in JavaScript when it is passed to a function with a return statement, without using built-in functions (.reverse()<

Answer 1

var str = 'sample string';
[].map.call(str, function(x) {
  return x;
}).reverse().join('');

OR

var str = 'sample string';
console.log(str.split('').reverse().join(''));

// Output: 'gnirts elpmas'

Answer 2

In ECMAScript 6, you can reverse a string even faster without using .split('') split method, with the spread operator like so:

var str = [...'racecar'].reverse().join('');

Answer 3

//es6
//array.from
const reverseString = (string) => Array.from(string).reduce((a, e) => e + a);
//split
const reverseString = (string) => string.split('').reduce((a, e) => e + a); 

//split problem
"                                                                    

                                                        
            

Answer 4

Detailed analysis and ten different ways to reverse a string and their performance details.

http://eddmann.com/posts/ten-ways-to-reverse-a-string-in-javascript/

Perfomance of these implementations:

Best performing implementation(s) per browser

• Chrome 15 - Implemations 1 and 6
• Firefox 7 - Implementation 6
• IE 9 - Implementation 4
• Opera 12 - Implementation 9

Here are those implementations:

Implementation 1:

function reverse(s) {
  var o = '';
  for (var i = s.length - 1; i >= 0; i--)
    o += s[i];
  return o;
}

Implementation 2:

function reverse(s) {
  var o = [];
  for (var i = s.length - 1, j = 0; i >= 0; i--, j++)
    o[j] = s[i];
  return o.join('');
}

Implementation 3:

function reverse(s) {
  var o = [];
  for (var i = 0, len = s.length; i <= len; i++)
    o.push(s.charAt(len - i));
  return o.join('');
}

Implementation 4:

function reverse(s) {
  return s.split('').reverse().join('');
}

Implementation 5:

function reverse(s) {
  var i = s.length,
      o = '';
  while (i > 0) {
    o += s.substring(i - 1, i);
    i--;
  }
  return o;
}

Implementation 6:

function reverse(s) {
  for (var i = s.length - 1, o = ''; i >= 0; o += s[i--]) { }
  return o;
}

Implementation 7:

function reverse(s) {
  return (s === '') ? '' : reverse(s.substr(1)) + s.charAt(0);
}

Implementation 8:

function reverse(s) {
  function rev(s, len, o) {
    return (len === 0) ? o : rev(s, --len, (o += s[len]));
  };
  return rev(s, s.length, '');
}

Implementation 9:

function reverse(s) {
  s = s.split('');
  var len = s.length,
      halfIndex = Math.floor(len / 2) - 1,
      tmp;


     for (var i = 0; i <= halfIndex; i++) {
        tmp = s[len - i - 1];
        s[len - i - 1] = s[i];
        s[i] = tmp;
      }
      return s.join('');
    }

Implementation 10

function reverse(s) {
  if (s.length < 2)
    return s;
  var halfIndex = Math.ceil(s.length / 2);
  return reverse(s.substr(halfIndex)) +
         reverse(s.substr(0, halfIndex));
}

Answer 5

var str = "my name is saurabh ";
var empStr='',finalString='';
var chunk=[];
function reverse(str){
var i,j=0,n=str.length;
    for(i=0;i<n;++i){
        if(str[i]===' '){
            chunk[j]=empStr;
            empStr = '';
            j++;
        }else{
            empStr=empStr+str[i];
        }
    }
    for(var z=chunk.length-1;z>=0;z--){
        finalString = finalString +' '+ chunk[z];
        console.log(finalString);
    }
    return true;
}
reverse(str);

Answer 6

function reverseString(string) {
    var reversedString = "";
    var stringLength = string.length - 1;
    for (var i = stringLength; i >= 0; i--) {
        reversedString += string[i];
    }
    return reversedString;
}