Yes. In the first diagram ("From the web"), D5 protects the regulator chip from the situation where the OUT terminal become positive with respect to the IN terminal. This happens when the input voltage disappears (i.e. pin 1 goes to GND), but the charge in capacitors C5 and C7 maintains a positive voltage on the OUT terminal. This can damage a 78-series regulator. D5 allows C5 and C7 to discharge via the (now at 0V) input wire.
D7 protects the circuit connected the output from reverse polarity. If the input polarity is the wrong way round, so GND is positive with respect to input, then D7 and D5 put a near-short circuit across the supply, blowing the fuse further back up the line, protecting the circuit connected to the output.
In "my design', D1 protects the regulator and the circuit connected to the output from reverse polarity of the input. However, this comes at the cost of dropping 0.6V or more across the diode in normal operation. In this circuit, switching off the input cannot pull the IN terminal of the regulator down to GND potential because D1 prevents that, and the charge on C2 holds the IN terminal positive until the circuit connected to the output has discharged C3 and C4.
So "my design" does the same job as "From the web", but with one less diode, at the cost of 0.6V-odd dropped across D1 in normal use.