A linear regulator will be no different to a resistor - 50% efficiency maximum.
It's funny how I know these things but I haven't been taught them and aren't always good enough at maths to work them out. I often don't know how I figure it out so I can't show my workings. This often used to be my downfall at college, some other people in my class seemed to be able to do the maths but didn't have the understanding, then there were the real geniuses who knew both.
Fortunately a constant current source is simpler so I can calculate it. Take a 1F capacitor, charged to 1V using a perfect 1A constant current linear regulator connected to a 1V supply:
[latex]t =\frac{CV}{I} = 1s\\
E_{IN} = VIt = 1 \times 1 \times 1 = 1J\\
E_(C) = \frac{1}{2}CV^2= \frac{1}{2}=1\times 1^2 = 0.5J[/latex]
This is assuming it's a perfect linear regulator with a dropout of 0V and a perfect capacitor, in reality it'll be worse than a resistor.