$7 \cdot 86$ Let $S=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ be a permutation of the natural numbers $1,2, \cdots, n$. $f(S)$ is defined as the minimum value of the absolute differences between every two adjacent elements in $S$. Find the maximum value of $f(S)$.
[Solution] The maximum value of $f(S)$ is $\left[\frac{n}{2}\right]$. We will prove this by considering two cases.
(1) When $n=2 k$, the absolute difference between $k$ and its adjacent numbers is no greater than $k$, so $f(S) \leqslant k=\left[\frac{n}{2}\right]$. On the other hand, let $S=(k+1,1, k+2,2, \cdots, 2 k, k)$, then $f(S)=k$.
(2) When $n=2 k+1$, the absolute difference between $k+1$ and its adjacent numbers is no greater than $k$, thus $f(S) \leqslant k=\left[\frac{n}{2}\right]$. In the permutation $S=(k+1,1, k+2,2, \cdots, 2 k, k, 2 k+1)$, $f(S)=k$, so we still have $f(S)=k=\left[\frac{n}{2}\right]$.