Example 2 Someone wants to go upstairs, and this person can go up 1 step or 2 steps at a time. If one floor has 18 steps, how many different ways can he go up one floor?
Solution 1: This person can climb one floor in a maximum of 18 steps and a minimum of 9 steps. Here, $a_{18}, a_{17}, a_{16}, \cdots, a_{9}$ represent the number of ways this person can climb the stairs in 18 steps, 17 steps, 16 steps, $\cdots, 9$ steps, respectively (for any two consecutive step counts, the latter takes 2 fewer 1-step moves and 1 more 2-step move, resulting in 1 fewer step). Considering the 2-step moves in each step, we easily get
$$
a_{18}=\mathrm{C}_{18}^{1}, a_{17}=\mathrm{C}_{17}^{1}, a_{16}=\mathrm{C}_{16}^{2}, \cdots, a_{9}=\mathrm{C}_{9}^{9} .
$$

In summary, the number of ways he can climb one floor is
$$
\begin{array}{l}
\mathrm{C}_{18}^{0}+\mathrm{C}_{17}^{1}+\mathrm{C}_{16}^{2}+\cdots+\mathrm{C}_{9}^{9} \\
=1+17+120+\cdots+1=4181
\end{array}
$$

Solution 2: Let $F_{n}$ represent the number of ways to climb $n$ steps.

Obviously, $F_{1}=1, F_{2}=2$. For $F_{3}, F_{4}, \cdots$, there are only two different ways to start: climb 1 step or climb 2 steps.

Therefore, for $F_{3}$, if the first step is 1 step, there are 3-1 $=2$ steps left, which counts as $F_{2}$ different ways; if the first step is 2 steps, there are 3-2 =1 step left, which counts as $F_{1}$ different ways. In total, $F_{3}=F_{2}+F_{1}=2+1=3$.
$$
\begin{array}{l}
\text { Similarly, } F_{4}=F_{3}+F_{2}=3+2=5, \\
F_{5}=F_{4}+F_{3}=5+3=8, \\
\cdots \cdots \\
F_{18}=F_{17}+F_{16}=2584+1597=4181 .
\end{array}
$$