Example 3 (1994 Asia Pacific Mathematical Olympiad Problem) Let $f$ be a function from $\mathbf{R} \rightarrow \mathbf{R}$, and
(1) For any $x, y \in \mathbf{R}, f(x)+f(y)+1 \geqslant f(x+y) \geqslant f(x)+f(y)$;
(2) For any $x \in[0,1)$, $f(0) \geqslant f(x)$;
(3) $-f(-1)=f(1)=1$.

Find all functions $f$ that satisfy the conditions.
From (1) and (2), we have $2 f(0)+1 \geqslant 2 f\left(\frac{1}{2}\right)+1 \geqslant f(1) \geqslant f(1)+f(0)$,
but from (3) we know $f(1)=1$, so the above inequality can be simplified to $2 f(0) \geqslant 0 \geqslant f(0)$, thus $f(0) \geqslant 0 \geqslant f(0)$, which implies $f(0)=0$.
From (2), for any $x \in(0,1)$, we have $f(x) \leqslant 0$.
If there exists $x \in(0,1)$ such that $f(x)f(x)+f(1-x)+1 \geqslant f(1)=1$,
$$

this leads to a contradiction. Therefore, for any $x \in[0,1)$, we have $f(x)=0$.
From (1), for any $x \in \mathbf{R}$, we have $f(x+1) \geqslant f(x)+f(1)=f(x)+1$.
Also, from (1) and (3), for any $x \in \mathbf{R}$, we have
$$
f(x) \geqslant f(x+1)+f(-1)=f(x+1)-1 .
$$

From the above two inequalities, for any $x \in \mathbf{R}$, we have $f(x+1)=f(x)+1$.
From (1) and (2), for any $x \in \mathbf{R}$, we have $f(x)=[x]$ (where $[x]$ denotes the greatest integer not exceeding $x$).