I want to reverse string

"My! name.is@rin"

My output must be

"yM!  eman.si@nir"

I wrote following code which gives output as "yM eman si nir" Please resolve the issue :

return str = stringIn.split("").reverse().join("").split(/[^a-zA-Z]/g).reverse().join(" ");

I want to reverse string

"My! name.is@rin"

My output must be

"yM!  eman.si@nir"

I wrote following code which gives output as "yM eman si nir" Please resolve the issue :

return str = stringIn.split("").reverse().join("").split(/[^a-zA-Z]/g).reverse().join(" ");

• javascript
• regex

• What do you think your code does? – Sverri M. Olsen Commented Oct 16, 2015 at 7:52
• Is it words you want to reverse, or the entire string? – OliverRadini Commented Oct 16, 2015 at 7:55
• try reversing each work + escape and unescape each word – Praveen Commented Oct 16, 2015 at 7:56

7 Answers 7

From your regular expression, it seems your criterion for "special characters" is anything other than the letters A to Z.

You can use String.prototype.replace with a regular expression to match sequences of letters, then provide a replace function that modifies the match before replacing it, e.g.

var stringIn = 'My! name.is@rin';
var rev = stringIn.replace(/[a-z]+/gi, function(s){return s.split('').reverse().join('')});
document.write(rev); // yM! eman.si@nir

Here is one more approach we can do the same:

function isLetter(char){
    return ( (char >= 'A' &&  char <= 'Z') ||
             (char >= 'a' &&  char <= 'z') );
}    

function reverseSpecialString(string){
    let str = string.split('');
    let i = 0;
    let j = str.length-1;
    while(i<j){
        if(!isLetter(str[i])){
            ++i;
        }
        if(!isLetter(str[j])){
            --j;
        }
        if(isLetter(str[i]) && isLetter(str[j])){
            var tempChar = str[i];
            str[i] = str[j];
            str[j] = tempChar;
            ++i;
            --j;
        }
    }
    return str.join('');
}

document.write(reverseSpecialString("Ab,c,de!$"));

let check=(str)=>{
    let pat=/[a-zA-z]/igm;
    if(str.match(pat)){
        return true
    }else{
        return false
    }
}

let r=(str)=>{
    let arr=str.split('');
    let l=0;
    let r=arr.length-1;
    while(l<r){
        if (!check(arr[l])){
            l++;
        }else if(!check(arr[r])){
            r--;
        }else{
            let tmp=arr[l];
            arr[l]=arr[r];
            arr[r]=tmp;
            l++;
            r--;
        }
    }
 return arr.join('');
}

document.write(r("p@$$word"));

Pythonic approach which answers in a single line.

import re

def reverseOnlyString(text_in):
    return ("".join([tup[0][::-1] + tup[1] for tup in re.findall(r'(\w*)(\W*)', text_in)]))

my_string = 'my name@#$ is 123 Test'
print(reverseOnlyString(my_string))

my_string = "My name is Tom Hank"
print(reverseOnlyString(my_string))
my_string = "MynameisTom Hank"
print(reverseOnlyString(my_string))
my_string = "%$%^^ *&&^ABC5443&***("
print(reverseOnlyString(my_string))

==>ym eman@#$ si 321 eliN
==>yM eman si moT knaH
==>moTsiemanyM knaH
==>%$%^^ *&&^3445CBA&***(

Pythonic approach without regex:

p = "a!!!b.c.d,e'f,ghi"
q = ''.join([i for i in p if i.isalpha() ])[::-1]


r = ''
for i in range(len(p)):
    if p[i].isalpha():
        r+=q[0]
        q =q[1::]
    else:
        r+=p[i]
print(r)
print(p)

Ruby Solution:

[46] pry(main)> "My! name.is@rin"
=> "My! name.is@rin"
[47] pry(main)> "My! name.is@rin".gsub(/\w+/) { |word| word.reverse }
=> "yM! eman.si@nir"

def reverse(str1):
    l1=list(str1)
    n=len(str1)
    l=0
    r=n-1
    while l<r:
        if not l1[l].isalpha():
            l+=1
        elif not l1[r].isalpha():
            r-=1
        else:
            l1[l],l1[r]=l1[r],l1[l]
        l+=1
        r-=1
    return ''.join(l1)

str1="a,b$c"
q=reverse(str1)
print(q)

c,b$a

uses a simple approach that traverses the string twice. if a special character is encountered then just proceed . else swap the alphanumeric character. the check for alphanumeric character is done using x.isalpha()