Solution:
Let $\alpha$ be the unique positive root of $x^{2018} - 11x - 24 = 0$.

So,
$$\alpha^{2018} = 11\alpha + 24$$
We are to find $\left\lfloor \alpha^{2018} \right\rfloor$.

Let us estimate $\alpha$.

Suppose $\alpha > 1$. Try $\alpha = 2$:
$$2^{2018} - 11 \times 2 - 24 = 2^{2018} - 46$$
which is very large and positive. Try $\alpha = 1$:
$$1^{2018} - 11 \times 1 - 24 = 1 - 11 - 24 = -34$$
So the root $\alpha$ is between $1$ and $2$.

Let us try $\alpha = 1.01$:
$$(1.01)^{2018} \approx e^{2018 \times 0.01} = e^{20.18} \approx 5.8 \times 10^8$$
So $\alpha$ is much closer to $1$ than $2$.

But let's use the equation:
$$\alpha^{2018} = 11\alpha + 24$$
Let $A = \alpha^{2018}$. Then $A = 11\alpha + 24$.

But $\alpha$ is just slightly more than $1$, so $\alpha = 1 + \varepsilon$ for small $\varepsilon > 0$.

Then,
$$\alpha^{2018} = (1 + \varepsilon)^{2018} \approx 1 + 2018\varepsilon$$
So,
$$1 + 2018\varepsilon = 11(1 + \varepsilon) + 24 = 35 + 11\varepsilon$$
So,
$$1 + 2018\varepsilon \approx 35 + 11\varepsilon$$
$$2018\varepsilon - 11\varepsilon \approx 34$$
$$2007\varepsilon \approx 34$$
$$\varepsilon \approx \frac{34}{2007}$$
So,
$$\alpha \approx 1 + \frac{34}{2007}$$
Now,
$$\alpha^{2018} \approx (1 + \frac{34}{2007})^{2018}$$
Let $x = \frac{34}{2007} \approx 0.01694$.

Then,
$$(1 + x)^{2018} \approx e^{2018x} = e^{2018 \times 0.01694} = e^{34.2} \approx 6.6 \times 10^{14}$$
But let's use the equation $\alpha^{2018} = 11\alpha + 24$.

So,
$$\alpha^{2018} = 11\alpha + 24 \approx 11 \times (1 + \frac{34}{2007}) + 24 = 11 + 11 \times \frac{34}{2007} + 24 = 35 + \frac{374}{2007}$$
Now,
$$\frac{374}{2007} \approx 0.186$$
So,
$$\alpha^{2018} \approx 35.186$$
Therefore,
$$\left\lfloor \alpha^{2018} \right\rfloor = 35$$