# LeetCode -- Unique Paths

When there’re two methods, I’ll always pick the easier ones.

This is a traditional combinatorial problem. It’s equivalent to, given `m-1`

black blocks and `n-1`

red blocks, the total number of combinations within.

It’s equivalent because there are (m-1) downs, and (n-1) rights to go; one can go in any order.

Java’s BigInteger is used to avoid overflow problem.

A factorial lookup table is pre-computed to avoid duplicate calculation.