One question about Regular Expression!!!

Question

I need to creat such a regular expression to match the format "[ ][ ][ ]".
For example, there is a context,
(1), " The project manager defines [1][0.400][+goals] for iterations."
Suppose that there are some spaces or "\n" characters in this way,
(2), " The project manager defines [    1 ] [  0.400   ]
[   +goals] for iterations."
If the pattern match the format succefully, (2) strings should be replaced by (1)strings, in order words, the format of (1) is what I need finally,
I had ever tried creating a regular expression likes \\[([^\n\s]]+)\\]\\[([^\n\s]]+)\\]\\[([^\n\s]]+)\\] , but it does not work well!
DO YOU HOW TO IMPLEMENT IT IN JAVA?
Thanks for your any reply!

Answer 1

What I really need is that, via the regular
expression, all the spaces and \n characters in
square brackets [ and ], ] and [, will be thrown
away.
For example,
Original:
1) "The project manager defines [   1  ] [
0.400 ]
[   +goals] for iterations with the support"
After matching:
2) "The project manager defines [1][0.400][ [+goals]
for iterations with the support"
String 2) is what I need finally!
Thanks for your any reply!Well I gave you the answer to that one already :-)
If you need to preserve the spaces in between words use this one. I'm sure there's a better way to do it, I'm no RegEx master.
    public static void main(String[] args)
        String s = "[ 1 ] [ 0.400 ]\n[ +go als]";
        System.out.println( "Before: " + s );
        System.out.println( "\n\n" );
        s = s.replaceAll( "\\[\\s+", "[" );
        s = s.replaceAll( "\\s+\\]", "]" );
        s = s.replaceAll( "\\]\\s+\\[", "][" );
        System.out.println( "After: " + s );
    }