Recursion: powerset , returning an array list of subsets

Question

Hi, i've checked hundreds of webpages, but i still can;t figure out how to create a recursive method that returns the powersets of a set.
The sets returned are of type Arraylists
ex: set {1, 2, 3} returns an arrayList of arrayLists: {   { }, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }.
The code should look somewhat like that: but the recursive call does not work...
ArrayList powerSet(ArrayList s) {
ArrayList result = new ArrayList();
if (s.size() == 1) { // base case
// return a new ArrayList containing just s
ArrayList tempAL = new ArrayList();
tempAL.add(s);
result.add(tempAL);
return result;
else {
Integer last = (Integer)s.get(s.size()-1); // separate the first character from the rest
ArrayList rest = new ArrayList(); // remainder of the set
rest.addAll(s);
rest.remove(rest.size()-1); // get rid off element "last"
// recursively call method on the remainder of the set
ArrayList L = new ArrayList();
L.add( powerSet(rest) ); // recursively call method on the remainder of the set
// return the set composed of the Union of
// the powerset of the set without the element (from the recursive call)
result.addAll(L);
// return the same set but
// with each element therein unioned with the element initially taken out
for(int i=0; i< L.size(); i++) {
ArrayList tmp = new ArrayList();
tmp.addAll( (ArrayList)L.get(i) );
tmp.add(last); // adding the element initially taken out
result.add(tmp);
return result;
Please someone help me. I know there has to be many errors during recursion. it's like i can't grasp the concept.
Thx a lot :)
Edited by: jonaphin on Oct 18, 2007 12:01 AM
btw, the output is: arraylists of arraylists of arraylists, depending on the depth of the recursion. argh :(
[[[[[[1]]], [[[1]], 2]], [[[[1]]], [[[1]], 2], 3]], [[[[[1]]], [[[1]], 2]], [[[[1]]], [[[1]], 2], 3], 4]]

Answer 1

I think he changed
if (s.size() == 1) { // base case
to
if (s.size() == 0) { // base case
Also, jonaphin are you in Alexander Pelin's Data Structure Class at FIU? Because, amazingly enough... this is the exact assignment that has to get handed in today! (October 18) lol ;)