If you only want to compute homology of a finite presentation complex $X$ of a group $G$ (as I understand from comments), you can do it by (integer) linear algebra only.
Let generators be $x_1,\dots,x_m$, with relations $r_1,\dots,r_n$.
Then $H_1(X)=G_{ab}$ is the cokernel of the $m\times n$ integer matrix $A$ of degrees $\deg_{x_i} r_j$ (view as morphism $\mathbb{Z}^n\to \mathbb{Z}^m$), and $H_2(X)$ is the kernel of $A$ (in $\mathbb{Z}^n$).
Note that the computation of $H_2$ of the group $G$ from a finite presentation is algorithmically infeasible in general, as shown by C. Gordon in the eighties (see this answerthis answer).
I like to see this problem as akin to that of deciding finite sets of tiles that tile the plane, as it asks for computing all tilings (or at least a "generating" set of tilings) of the 2-sphere by copies of topological disks boundary-decorated with relations (well, maybe introducing some non-reduced relator words). But I don't know if this idea leads to a proof (Gordon's proof is different). Recall that deciding finite sets of tiles that tile the plane is also algorithmically infeasible.
When possible, this computes $H_2(G)$ of the group as $H_2(X)/h(\pi_2(X))$, where $h$ is Hurewicz homomorphism.