Hi,
A quad of quick points...
First, R2 is there to prevent a surge when the unit is first plugged in. 18 ohms only limits the current surge to around 30 amps while 100 ohms comes in around 6 amps. Of course 100 ohms uses more power and requires a larger resistor, so the choice is yours. There are also PTC thermistors used for this purpose that have resistance that gets lower after the circuit operates for a while. You may want to think about this because this is not your typical 20ma off line LED circuit.
Second, a full wave bridge is used for better utilization of the capacitor. The capacitor is probably the most expensive part in the circuit, so you want to get all you can out of it. What this means is that for a full wave bridge that drives the LED at around 200ma and uses 3uf, it will require DOUBLE that value (6uf) for half wave operation to get the same 200ma average current. Using two more cheap diodes (you need a second diode anyway for half wave) is a small price to pay for two times the current.
Third, the 3uf cap should be rated for AC operation, and its rating should be no less than 1.5 times the voltage of the line, or better yet 2 times.
Fourth, the electrolytic has to have enough capacitance to maintain the voltage the LED needs, but with no resistor between that cap and the LED this probably wont happen anyway. The LED limits the voltage the cap can charge up to, and with the cap being only 220uf it can easily discharge quickly below the LED voltage, meaning it probably wont do anything much. We can look into this more. The better design uses a series resistor so that the cap can have some flexing room while the LED draws current between half cycles. True, the resistor uses some power too, so a tradeoff is required between a really good value and a value that works more or less ok.