In this paper we revisit the classical regular expression matching problem, namely, given a regular expression $R$ and a string $Q$, decide if $Q$ matches one of the strings specified by $R$. Let $m$ and $n$ be the length of $R$ and $Q$, respectively. On a standard unit-cost RAM with word length $w \geq \log n$, we show that the problem can be solved in $O(m)$ space with the following running times: \begin{equation*} \begin{cases} O(n\frac{m \log w}{w} + m \log w) & \text{if $m > w$} \\ O(n\log m + m\log m) & \text{if $\sqrt{w} < m \leq w$} \\ O(\min(n+ m^2, n\log m + m\log m)) & \text{if $m \leq \sqrt{w}$.} \end{cases} \end{equation*} This improves the best known time bound among algorithms using $O(m)$ space. Whenever $w \geq \log^2 n$ it improves all known time bounds regardless of how much space is used.