Check for repeated characters in a string Javascript

Solution 1:

(A recursive solution can be found at the end, of this answer.)

You could use javascript builtin Array functions some MDN some reference

 var text = "test".split("");
 text.some(function(v,i,a){
   return a.lastIndexOf(v)!=i;
 });

callback parameters:
v ... current value of the iteration
i ... current index of the iteration
a ... current array

.split("") create an array from a string
.some(function(v,i,a){ ... }) goes through an array until the function returns true, and ends than right away. (doesn't loop through the whole array, if it finds an match earlier)

Details to the some function can be found here

Tests, with several strings:

var texts = ["test", "rest", "why", "puss"];

for(var idx in texts){
    var text = texts[idx].split("");
    document.write(text + " -> " + text.some(function(v,i,a){return a.lastIndexOf(v)!=i;}) +"<br/>");
    
  }
  //tested on win7 in chrome 46+

If recursion is needed.

Update for recursion:

//recursive function
function checkString(text,index){
    if((text.length - index)==0 ){ //stop condition
        return false; 
    }else{
        return checkString(text,index + 1) 
        || text.substr(0, index).indexOf(text[index])!=-1;
    }
}

// example Data to test
var texts = ["test", "rest", "why", "puss"];

for(var idx in texts){
    var txt = texts[idx];
    document.write( txt +  " ->" + checkString(txt,0) + "<br/>");
}
//tested on win7 in chrome 46+

Solution 2:

This will do:

function isIsogram (str) {
    return !/(.).*\1/.test(str);
}

Solution 3:

you can use .indexOf() and .lastIndexOf() to determine if an index is repeated. Meaning, if the first occurrence of the character is also the last occurrence, then you know it doesn't repeat. If not true, then it does repeat.

var example = 'hello';

var charRepeats = function(str) {
    for (var i=0; i<str.length; i++) {
      if ( str.indexOf(str[i]) !== str.lastIndexOf(str[i]) ) {
        return false; // repeats
      }
    }
  return true;
}

console.log( charRepeats(example) ); // 'false', because when it hits 'l', the indexOf and lastIndexOf are not the same.