Look at the circuit I posted earlier, with the NOR gates.
Look at how each gate has two inputs, going to the outputs of the other gates. To design the same circuit, but with four switches, use three input NAND gates, again with each input going to the output of the other gates.
Here's a discrete version, showing it can be done without ICs. It works on the same principle, but using discrete NOR gates. Rather than shorting the output to +V to force the output high, the switch shorts the base to 0V. Put C1 across the switch you want it to default to, when the power is first applied.
It's possible to build a simlar circuit using NAND gates, made with Darlington pairs/MOSFETs, which can drive the loads directly, but it's only suitable for high supply voltages.EDIT:
It's easy. Rather than just resistors, use diodes, as well as resistors, to make AND gates before the MOSFETs. If the switches can take the 700mA, required by the LEDs, then the 1k resistors can be eliminated (replaced with a short) and the switches put in parallel with the MOSFETs (drain to source). If the LEDs glowing dimly and associated voltage drop, when they're off isn't an issue, the 10k resistors can be eliminated (replaced with an open circuit)
This circuit will run directly from the constant current source, as long as the forward voltage of the LEDs is high enough MOSFETs on enough. Note, the N=2 next to LED, means that each LED on the schematic represents two models in series, giving a voltage drop of about 7V. Again C1 is in parallel with the switch, which the circuit defaults to, when power is first applied.