Does voltage or current affect the time to heat a bed more?
It depends on whether you are re-using the bed or not, it is actually the resistance of the bed that determines this in conjunction with the voltage (the current you get for free).

Let's say that the heatbed resistance is 1.2&nbsp;&ohm; (depending on the heated bed make and model the resistance is typically in between 0.9&nbsp;-&nbsp;1.5&nbsp;&ohm;), this means that the power can be calculated using:
$$P = U \times I$$
$$U = I \times R$$
combining gives:
$$ P = I^2\times R = \frac{U^2}{R} $$

For 12&nbsp;V (assumed default printer voltage) this means that the heatbed power equals about 120&nbsp;Watt (at a current of 10&nbsp;A). Running that same bed at 24&nbsp;V means that the power is 480&nbsp;Watt (at a current of 20&nbsp;A). So yes, that will heat up fast, at the expense of an increased current, which is pretty high, and should not be attempted without extra resistance in the loop.

If you're using the laptop charger, the current draw equals about 16&nbsp;A, which the adapter cannot deliver.

This means that you need to acquire a new heatbed that is able to handle a higher voltage out of the box (more resistance), or you need to put additional resistors in the loop, but beware of the currents. Note that heated beds for 12&nbsp;V/24&nbsp;V exist, the wiring is different depending on the voltage. Note that such beds heat up faster, it all depends on the resistance and the voltage, but running the 24&nbsp;V circuit on 19.5&nbsp;V (160&nbsp;Watt bed) is definitely an improvement over the 120&nbsp;Watt bed at 12&nbsp;V but still requires about 8&nbsp;A (only applicable to heatbed that can run 12&nbsp;V/24&nbsp;V through extra resistance connections).

Be careful with this and be sure what you are doing!