Just one simple? regex :)

#!/usr/bin/perl

# https://perlmonks.org/?node_id=1230512

use strict;
use warnings;

$_ = 117;
print "start at $_\n";

s/.*\K # find the last
  (.) # digit such that
  (.*) # there is a later (latest)
  (.)(??{$1 >= $3 and 'x'}) # digit greater than it
  (.*) # and get rest
  # swap those two digits ( $1 & $3 )
  # then reverse everything after the first swapped digit
  / $3 . reverse $2.$1.$4 /xe;

print " next is $_\n";
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Outputs:

start at 117
 next is 171
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