Delete duplicate elements from an array [duplicate]

As elements are yet ordered, you don't have to build a map, there's a fast solution :

var newarr = [arr[0]];
for (var i=1; i<arr.length; i++) {
   if (arr[i]!=arr[i-1]) newarr.push(arr[i]);
}

If your array weren't sorted, you would use a map :

var newarr = (function(arr){
  var m = {}, newarr = []
  for (var i=0; i<arr.length; i++) {
    var v = arr[i];
    if (!m[v]) {
      newarr.push(v);
      m[v]=true;
    }
  }
  return newarr;
})(arr);

Note that this is, by far, much faster than the accepted answer.

var arr = [1,2,2,3,4,5,5,5,6,7,7,8,9,10,10];

function squash(arr){
    var tmp = [];
    for(var i = 0; i < arr.length; i++){
        if(tmp.indexOf(arr[i]) == -1){
        tmp.push(arr[i]);
        }
    }
    return tmp;
}

console.log(squash(arr));

Working Example http://jsfiddle.net/7Utn7/

Compatibility for indexOf on old browsers

you may try like this using jquery

 var arr = [1,2,2,3,4,5,5,5,6,7,7,8,9,10,10];
    var uniqueVals = [];
    $.each(arr, function(i, el){
        if($.inArray(el, uniqueVals) === -1) uniqueVals.push(el);
    });

Try following from Removing duplicates from an Array(simple):

Array.prototype.removeDuplicates = function (){
  var temp=new Array();
  this.sort();
  for(i=0;i<this.length;i++){
    if(this[i]==this[i+1]) {continue}
    temp[temp.length]=this[i];
  }
  return temp;
} 

Edit:

This code doesn't need sort:

Array.prototype.removeDuplicates = function (){
  var temp=new Array();
  label:for(i=0;i<this.length;i++){
        for(var j=0; j<temp.length;j++ ){//check duplicates
            if(temp[j]==this[i])//skip if already present 
               continue label;      
        }
        temp[temp.length] = this[i];
  }
  return temp;
 } 

(But not a tested code!)