I believe that:

s/(?<=[[:alpha:]])\s+(?=[[:alpha:]])/ /g;
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Will have the same effect as your posted expression. Although it will replace a single space character between two letters with a single space character (essentially a no-op), because it is a simpler pattern will actually take less time to execute. Below I am posting a simple benchmark I used to test this theory. If I have missed something please enlighten me.

#!/usr/local/bin/perl -w
use strict;
use Benchmark;

my(@tests) = <DATA>;

timethese(400, {his => sub{ s/(?<=[[:alpha:]])(?:\s\s+|[^\S ]+)(?=[[:a
+lpha:]])/ /g foreach(@tests); },
                mine => sub{s/(?<=[[:alpha:]])\s+(?=[[:alpha:]])/ /g f
+oreach(@tests); }});


__DATA__
    A sting             with weird
A string without weird
                Another variety with            more wierd
Anotherthingwithnospaces
        something odd
soemthing normal


[me@mylinux]$ ./space.pl
Benchmark: timing 400 iterations of his, mine...
       his:  0 wallclock secs ( 0.07 usr +  0.00 sys =  0.07 CPU) @ 57
+14.29/s (n=400)
            (warning: too few iterations for a reliable count)
      mine:  0 wallclock secs ( 0.03 usr +  0.00 sys =  0.03 CPU) @ 13
+333.33/s (n=400)
            (warning: too few iterations for a reliable count)
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