Example 7 A shipping company has a ship leaving Harvard for New York every noon, and at the same time every day, a ship also leaves New York for Harvard. It takes seven days and seven nights for the ships to complete their journeys in both directions, and they all sail on the same route. How many ships of the same company will the ship leaving Harvard at noon today encounter on its way to New York?
Solution 1: Establishing Equal Relationships of Distance and Time.
Analysis: If this ship meets another ship opposite to it at point $A$, then the next ship opposite to it is at point $B$. By the equal relationship of distance, it is easy to know that they meet at the midpoint of $A B$, and the interval is exactly half a day. Therefore, throughout the process, it should meet 15 ships coming from the opposite direction.
Solution 2: Establishing Indeterminate Equation Method.
Let the ship that departs from Harvard at noon today meet a ship that departed $x$ days ago, then
\[
\begin{array}{l}
x+y=7(\hat{u} \leqslant x \leqslant 7), \\
x-y=n(n \in Z). \\
\therefore 2 x=7+n. \\
\text { Thus, }-7 \leqslant n \leqslant 7.
\end{array}
\]

Therefore, $n$ can take $-7,-6, \cdots,-1,0,1, \cdots, 6,7$. There are a total of 15 ships.