Easier and simpler, but less efficient

Woah! I would benchmark that before making such a statement. I expect you'll need a seriously huge list before the O(n log n) starts to lose to your straight O(n).

There's no debate that it will, but the problem is the 0(n) algorithm is using a comparatively large number of slow ops, where as the sort compiles down into a single op, and there you're running at C speed (simple {$a <=> $b} blocks are recognised and special-cased during the parse). This will drown out the extra cost for a long, long time (that is: for a long list of values). On my machine, the cross-over occurs between 100 000 and 1 000 000 elements (and I had to run the million element benchmark for 15 seconds in order to give it enough time to settle down)

               Rate with_scan_1 with_sort_1
with_scan_1 35662/s          --        -53%
with_sort_1 76332/s        114%          --
               Rate with_scan_2 with_sort_2
with_scan_2  6838/s          --        -40%
with_sort_2 11437/s         67%          --
             Rate with_scan_3 with_sort_3
with_scan_3 759/s          --        -11%
with_sort_3 853/s         12%          --
              Rate with_sort_4 with_scan_4
with_sort_4 62.6/s          --        -18%
with_scan_4 76.2/s         22%          --
              Rate with_sort_5 with_scan_5
with_sort_5 3.38/s          --        -47%
with_scan_5 6.38/s         89%          --
            s/iter with_sort_6 with_scan_6
with_sort_6   4.88          --        -66%
with_scan_6   1.65        196%          --
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And since either choice is crazy fast enough for me, I'd throw my lot in with the more succinct version -- less chance of introducing semantic mistakes and off-by-one errors). For instance, I had to think for a little while about how you initialised $min and $max...