There are several ways to solve this problem. I mean, you can use a simplicial, singular or cellular homolgy theory to solve this problem, resulting the same homolgy groups in this case.

The simplest one might be a simplicial homology using -complexes.

A circle S is described as one vertex v (0-simplex) and one edge (1-simplex) in the -complex structure. Since the boundary map of 1-simplex e is , we have , where since we have a single 0-simplex (vertex v) whose boundary map is trivially zero.

In a similar manner, we have since there is no 2-simplex in our -complex structure and we have a single 1-simplex (edge e) whose boundary map is zero.

It is trivial to show that all the remaining higher homology groups of S are zero.

This is kind of a standard textbook question and you can find this in most algebraic topology books like Hatcher's Algebraic topology p 106.