Example 4 Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $f(1)=1, f(2)=2, f(3)=3$, then the value of $\frac{1}{4}[f(4)+f(0)]$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8
Solution: Let $g(x)=f(x)-x$, then
$$
\begin{array}{l}
g(1)=g(2)=g(3)=0, \\
f(x)=g(x)+x
\end{array}
$$

Therefore, $g(x)=(x-1)(x-2)(x-3)(x-r)$.
( $r$ is a constant)
$$
\begin{aligned}
\therefore & \frac{1}{4}[f(4)+f(0)] \\
& =\frac{1}{4}[g(4)+4+g(0)] \\
& =\frac{1}{4}[6(4-r)+4+6 r]=7 .
\end{aligned}
$$

Hence, the answer is (C).