Topological/homotopical classification for 1-dim CW-complexes?

Well, 1-dimensional CW complexes are graphs, and every (connected) graph is homotopically equivalent to a wedge sum of circles. Hence they are classified by their $\pi_1$ (actually just their Euler characteristic is sufficient if they are finite).

For details see Hatcher - Algebraic Topology (http://www.math.cornell.edu/~hatcher/AT/AT.pdf), sections 1.A and 1.B.

In general, connected 1-D CW complexes, i.e. connected graphs, are classified up to homotopy by the abelian group rank of their first homology, equivalently the free group rank of their fundamental group. If the graph happens to be finite, or if its rank is finite, then the Euler characteristic works too.

To classify general 1-D CW complexes up to homotopy equivalence, one simply counts how many components there are for each possible value of the rank.

On the other hand, classifying connected finite graphs up to homeomorphism is a delicate combinatorial job. I do not believe there are any closed form expressions for the number of homeomorphism types of graphs of a given rank, say, although there may be generating functions.

Additional remarks: To respond to OP's clarification in the comments, one can reduce the graph homeomorphism classification to graph isomorphism classification in the following manner. This can be done for arbitrary graphs, but the statement is simpler for finite connected graphs which are not homeomorphic to the circle $S^1$ and I'll restrict to that situation. (Also, in my discussion graphs are not "simplicial complexes": an edge can have both ends attached to the same vertex, and two edges can have their respective ends attached to the same pair of vertices).

Consider a finite connected graph $G$ not homeomorphic to $S^1$. By removing the valence $2$ vertices of $G$ from the vertex set one gets another graph structure on $G$, which I like to think of as the "natural" graph structure on $G$ (this doesn't work if $G$ is a circle, or an infinite graph homeomorphic to $\mathbb{R}$, or an infinite graph with an isolated "end").

Then what one shows is the following:

• If two finite connected graphs $G,G'$ have no valence $2$ vertices then $G,G'$ are homeomorphic if and only if they are isomorphic as graphs.

As the OP says in his comment, the hard part is the "only if" direction. The main fact needed for the proof is that the valence of a point $p$ in a graph is a local topological invariant: the valence of $p$ is equal to the number of connected components of $U-p$ for all sufficiently small connected neighborhoods $U$ of $p$. Also, assuming $G$ has no valence $2$ vertices, the points of $G$ with valence $2$ are precisely the points in the interiors of edges, and the points with valence $\ne 2$ are precisely the vertices. Putting this together, if $G,G'$ have no valence $2$ vertices then any homeomorphism $f : G \mapsto G'$ restricts to a bijection of vertices, and a bijection of edges that preserves endpoints, and so $f$ is a graph isomorphism.

I like to use this discussion as one of the first applications of connectivity in my algebraic topology course.