I'm not very topology-savvy, and the best I could come up with was:

Identify two (2-dimensional) closed discs to be the upper and lower hemispheres of S^2, and then chop these up into four pieces each; then each sector can be identified with a 2-simplex, with the right orientations. This gives a Δ-complex with eight 2-simplices, twelve 1-simplices, and six 0-simplices (by my counting).

If I'm perfectly honest, this seems quite complicated for such a simple topological space. Even the torus has only six simplices in all. Am I missing some simpler identifying or 'gluing' procedure?

EDIT: Sorry, I'm being silly. I think there is a way to do it identifying 2-simplices (triangles) to closed discs. Can anyone just confirm that when you do this correctly, the homology groups should be $\displaystyle H_n(S^2) = \{0\}$ for $\displaystyle n \neq 2$, and $\displaystyle H_2(S^2) = \mathbb{Z}$ ?