The de Sitter Universe is described by the Friedmann equation with $k=0$, $\rho=0$ and a positive cosmological constant $\Lambda$:
$$\Big(\frac{\dot{a}}{a}\Big)^2=\frac{\Lambda c^2}{3}$$
The solution for the scale factor $a(t)$ is given by:
$$a(t) \sim e^{\pm\sqrt{\Lambda c^2/3}\ t}$$
Thus a Universe with a positive cosmological constant $\Lambda$ expands exponentially. I suppose the solution with a negative square root is unphysical.
I understand that it is possible to have an anti-de Sitter Universe with a negative cosmological constant $-\Lambda$.
I have heard people say that such a Universe would shrink exponentially but surely it should oscillate?
$$a(t) \sim e^{\pm i\sqrt{\Lambda c^2/3}\ t}$$
Correction
Andrew pointed out that the metric must be real so that $a(t) \sim e^{iwt}$ is not a legitimate solution.
Instead of the first Friedmann equation I can take the acceleration equation for a de Sitter Universe with positive cosmological constant $\Lambda$ given by:
$$\frac{\ddot{a}}{a}=\frac{\Lambda c^2}{3}$$
The solution is given by:
$$a(t) \sim \sinh{\sqrt{\Lambda c^2/3}\ t}$$
Let us consider an anti-de Sitter Universe, with a negative cosmological constant $-\Lambda$. The acceleration equation is given by:
$$\frac{\ddot{a}}{a}=-\frac{\Lambda c^2}{3}$$
The real solution to this equation is given by:
$$a(t) \sim \sin{\sqrt{\Lambda c^2/3}\ t}$$
Therefore the anti-de Sitter Universe collapses as expected - not exponentially but like a sine function.
By substituting into the first Friedmann equation one finds that for both anti-de Sitter and de Sitter Universes we have negative spatial curvature $k \sim -\Lambda c^2/3$ (again as pointed out by Andrew).