Disregarding the last line of your question, yes, you do get a lattice of rank $L$, and this lattice can be computed explicitly. A quick and dirty algorithm would use
>MR0525944 (80j:10031) Reviewed
Nikulin, V. V.
Integer symmetric bilinear forms and some of their geometric applications. (Russian)
Izv. Akad. Nauk SSSR Ser. Mat. 43 (1979), no. 1, 111–177, 238.

The thing is that the lattice in question is a finite index extension of the original lattice, say, $S$. Restricting to $S$ the Euclidean inner product, we make it a true nondegenerate lattice, and then it has but finitely many finite index extensions: they are enumerated by the isotropic subgroups of the finite group $\operatorname{discr}S$. Each has an explicit basis (made by rational vectors in the basis of $S$), and one can try them all one by one to see if these vectors are actually integral.