How to determine whether a planar graph can be generated by a degree sequence? 
For the following degree sequences of graphs,
$2,2,2,3,3,3,3,4,5,5\\1,1,1,1,2,2,2,2,3,3$
which can be the degree sequence of a simple undirected
graph that is connected and planar?

I think only the second one can be a valid sequence. Is there a general rule to judge whether a degree sequence can generate a planar graph?
 A: In general it will require some thought on whether a degree sequence is that of a planar graph. For example, see this question for some possible strategies, which include using Kuratowski's Theorem, or the well known edge bound $3n - 6$. You can also compute the average degree, which for a planar graph must be strictly less than 6. (See Planarity Criterion.) You'll notice a lot of the time we check to see if a graph violates any of the criterion.
Both sequences may represent a planar graph.
For $2,2,2,3,3,3,3,4,5,5$, you may notice several things. Since we're assuming $G$ is connected, $G$ cannot be a tree (there are no degree $1$ vertices) and thus has a cycle. However, $|E(G)| = 16 \le 3(10) - 6 = 24$, so no luck there (remember, we can only use this bound to prove a graph is non-planar.) Perhaps after playing around you might think that there is a planar graph, and you'd be right. Using the Havel--Hakimi Algorithm, stopping short when I obtained the degree sequence of the path on $8$ vertices, we find the following graph:

From this example we see that it's not necessarily the case that $G$ has a cycle but no triangle (which would have enabled us to use the improved bound of $2n-4$). Even then, $|E(G)| = 16 = 2(10) - 4$.
For $1,1,1,1,2,2,2,2,3,3$, you can quickly notice that this graph cannot contain $K_5$ or $K_{3,3}$ as a subdivision and thus must be planar. This uses Kuratowski's Theorem. It also shouldn't be too hard to find a tree with this degree sequence.
For the first sequence, you may immediately notice that $G$ cannot have a $K_5$ subdivision, however $K_{3,3}$ may require more reasoning. I.e., unlike the second sequence, we've shown the first sequence can represent a planar graph, we haven't shown that it must.
