1.108 Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$ and can randomly jump to one of the two adjacent vertices each time. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different ways can the frog jump from the start to the stop?
［Solution］From the figure, we can see that the frog cannot be at point $D$ after 1, 2, or 4 jumps. Therefore, the possible jumping methods for the frog are only the following two:
（1）The frog reaches point $D$ after 3 jumps, with a total of 2 different jumping methods;
（2）The frog stops after a total of 5 jumps. At this time, the number of jumping methods where the frog is not at point $D$ after 3 jumps is $2^{3}-2=6$. The number of different jumping methods for the last 2 jumps is 4. Therefore, this type of jumping method totals 24.
In summary, there are 26 different jumping methods that meet the requirements of the problem.