I don't like it being defined as $\vec{r} \times \vec{mv}$ as the angular nature is not obvious in that definition.
Suppose there's a single particle moving around. We choose an arbitrary origin. We define the angular momentum at time $t$ as $m|\vec{r(t)}|^2$ times its angular velocity. Angular velocity at time $t$ is defined as the vector perpendicular to both $\vec{v(t)}$ and $\vec{r(t)}$ (according to some conventional rule), and having the magnitude $\frac{d\theta}{dt}$, where $\theta (t)$ is the angular position of the particle at time $t$ in the plane of $\vec{r(t)}$ and $\vec{v(t)}$, with respect to the chosen origin.
So this defines it for a single particle. For a system of particles, we just sum up the angular momenta. The formula $\vec{r}\times \vec{mv}$ is arrived at as a means of calculating it. Is this definition equivalent to $\vec{r}\times \vec{mv}$? Can either of these definitions be used for any general problem?