Sorting array by even and odd numbers

Question

Using javascript sort() method, I am trying to do sorting a list but sorting have in a group of even numbers and odd numbers.

The code which I tried is work

Answer 1

The short of the shortest:

n.sort(function(a, b) {
    return a % 2 - b % 2 || a - b;
});

To make it work with negative numbers we can add Math.abs():

n.sort(function(a, b) {
    return Math.abs(a % 2) - Math.abs(b % 2) || a - b;
});

Or even more compact variant using bitwise AND:

n.sort(function(a, b) {
    return (a & 1) - (b & 1) || a - b;
});

Answer 2

My try:

fiddle

var n = [10,20,21,4,5,6,7,99,0,-12,12,13,-45,10,20,21,-13,4,5,6,7,99,0,12,13,10,-99,20,21,4,5,6,7,99,0,12,13,10,20,21,4,5,6,7,99,-15,-18,0,12,13];

function sort(a, b) {
    if (!(a % 2) && !(b % 2)) {
        return a > b ? 1 : -1;
    }
    if ((a % 2) && (b % 2)) {
        return a > b ? 1 : -1;
    }
    if ((a % 2) && !(b % 2)) {
        return 1;
    }
    return -1;
}
console.log(n.sort(sort));​

Answer 3

Change the code as follows:

n.sort(function(a,b){
 if (a % 2 != b % 2 ){
   return a%2;
  }else {
      return (a - b) > 0 ? 1 : -1; 
  }
});

Working sample is here.

Edit:

n.sort(function(a,b){
 if (a % 2 != b % 2 ){
     return a%2 == 0 ? -1 : 1; // this is the fix :)
  }else {
      return (a - b) > 0 ? 1 : -1; 
  }
});

Edit 2: I've modifed the code for negative numbers. See working sample.

n.sort(function(a,b){
 if (a % 2 != b % 2 ){
   return Math.abs(a)%2;
  }else {
      return a > b ? 1 : -1; 
  }
});