Getting all possible ordering of vector

Question

hello
I need to find all possible order of a set of vector I mean if I have the following:
0
1
2
all posssible would be:
0 1 2, 1 0 2, 2 1 0, 1 2 0, 2 0 1, 0 2 1
I have tried to use the swap but I didn't get all possible set
I have used the following code:
Code:
for( int x=0;x<vector.size();x++)     {            
for ( int w=0;w< vector.size();w++){
Collections.swap(vector, x, w);}}
as I said before the swap method didn't provide me the result I want
can somebody tell me how to do it please urgent

Answer 1

warnerja wrote:
rdkh wrote:
Sorry I had to wait a few days to do this for you. Here is a "very creative" solution:
... (code snipped)If by "very creative" you mean "incorrect" or at least "convoluted", then you're right.As I always ask questions, I watch for any chance to contribute to the forum. But as I don't have confidence and knowledge, I only post info I read in a book or google, and code that compiles/works correctly. As far as I can test, my code works correctly.
I was inspired by a "very creative" way to calculate "pi" using a circle in a square and random numbers. With modern processors, knowing where to start and end is sometimes enough. Let my CPU tear into the problem, keep guessing to get "pi" to infinite precision, let the computer keep guessing until the requested collection of Vectors is found. While not AI, I find elegence in this approach.
P.S.
if(!used.contains(rnd)) {
used.add(rnd);
v.add(rnd);What's the point of both "used" and "v" containing exactly the same set of values? One of these is redundant.
And have you heard about Random.nextInt(int n)?I just know very basic APIs (thats why I hang-out only on "New to Java" forum). As you told me: "Set<Integer> used = new HashSet<Integer>();" was not needed.
And a special "thank you" goes out to you for trying to do a lazy bum's homework for him/her. May you end up "working with" such a person.ok, but I warned that my code is "proof of concept" and I said I wrote it just for fun, and it works correctly.
Edited by: rdkh on Jan 11, 2010 4:58 PM