Hi there!
Look at my approach for solving this problem.
I`m using 1-dimensional representation of digital matrix. In example matrix M below:
       A11, A12, A13, A14
 (M) = A21, A22, A23, A24
       A31, A32, A33, A34
       A41, A42, A43, A44
transforms to the vector V:
 V = (A11, A12, A13, A14, A21, A22, A23, A24, A31, A32, A33, A34, A41, A42, A43, A44)
(array of integers in my C++ implementation).
Than point out the order in which elements of matrix M should appears in result:
(1)  (3)  (6)  (10) (2)  (5)  (9)  (13) (4)  (8)  (12) (15) (7)  (11) (14) (16)
A11, A12, A13, A14, A21, A22, A23, A24, A31, A32, A33, A34, A41, A42, A43, A44, where
values in brackets are elements serial numbers in result.

Onwards take a pencil and draw a curves between elements in desired order:
(1)  (3)  (6)  (10) (2)  (5)  (9)  (13) (4)  (8)  (12) (15) (7) (11) (14) (16)
A11, A12, A13, A14, A21, A22, A23, A24, A31, A32, A33, A34, A41, A42, A43, A44
 |___↑_|__↑_________↑ |  ↑ |            ↑ |
     |_|__|___________|  | |            | |
       |__|______________|_|____________| |
          |              |_|______________|
          |________________|                    ... and so on

In example above 6 elements ordered by curves connections. It should be observed that
it`s possible to fix sequence of actions using numeric relative offsets. See below:
Step  1) +4 (+4 is offset (in elements) of first element A11);
Step  2) -3 (-3 is offset of element, obtained in step 1);
Step  3) +7 (+7 is offset of element, obtained in previous step and so on);
Step  4) -3
Step  5) -3
Step  6) +10
Step  7) -3
Step  8) -3
Step  9) -3
Step 10) +10
Step 11) -3
Step 12) -3
Step 13) +7
Step 14) -3
Step 15) +4

Halves of steps sequence are simmetrical. Sequence contains two operation type: forward offset and backward offset (all offset are relative). Value of backward always equals to (side of matrix - 1). Value of forward offset different on difference steps. On first and last steps it`s equal to side size. On other steps it`s increased or decreased (above or below of middle steps) on value (side of matrix - 1).
Forward steps amount is always equals to 2 x (side-1). Amount of backwards steps are 1 on first forward step, 2 on second and so on toward the middle of iterations. After middle backwards steps amount decreased by 1 on each iterations except last. On last iterations it`s not backwards steps.

This algorithm has simple (not required complex data structures) implementation on various language.

Thank you for your attention and Good luck!

P.S. Sorry for my English =)

P.P.S. Graphical representation of algorithms steps is distorted because when message is sent any occurance of double, triple and so on spaces are replaced on one space character :(

Edited by author 13.09.2011 15:09