How to change the teleporters to only teleport to each other?

Just like @agent86, I decided to hijack the loop. My solution is slightly shorter and links all 3 teleporter near the player to the ones near the exit.

var player = map.getPlayer();
var entry = [], exit = [];

for (i = 0; i < teleportersAndTraps.length; i++) {
    var t = teleportersAndTraps[i];
    if (t.getType() == 'teleporter') {
        if (Math.abs(t.getX() - player.getX()) <= 2 &&
            Math.abs(t.getY() - player.getY()) <= 2)
            entry.push(t);
        if (Math.abs(t.getX() - (map.getWidth() - 10)) <= 2 &&
            Math.abs(t.getY() - (map.getHeight() - 5)) <= 2)
            exit.push(t);
    }
}

for (i = 0; i < 3; i++)
    entry[i].setTarget(exit[i]);

break;

Basically, I'm looking for any teleporter that has a x- or y-distance of 2 or less to the player or the exit.

What i did is link each teleporter in the list to the teleporter that comes after it in the list, and because of that all the teleporters form one big loop. all i did was just always enter the teleporter i just went through, and after a few teleports i got to the exit.

In my case these solutions did not convince me. I actually just didn't like the idea of linking directly to the exit, what I did was looking for each teleporter for one a step closer to the exit, that way, no matter the path you choose, you will always end up where you want (the exit) by using the teleports and without any risk of falling into the traps (unless I guess if you jump into the last row or column of teleports, I tried and it actually crashes if you the link is empty [edit: actually not, it does not crash, I realized it just takes a little bit of time to load it and there are still traps in the last rows/columns, it might not be the most efficient timewise]... but hey, it works! hahaha).

Here is my code if you want to look at it: https://gist.github.com/Untrusted-Game/5f78e8633962adc9d34e28ecebac1e4a