in reply to Re: Finding the max()/min()
in thread Finding the max()/min()

Your recursive version of max is broken for lists like (-1,0,1,2) and (-1,undef,1,2). i.e. the code assumes it is at the end of the list whenever $next==0 or $next==undef (which isn't true in general). Here's a snazzy (if not the most efficient) recursive version with a hat tip to Zaxo... 
sub max {
    my ($x, @xs) = @_;
    @xs ? ($x, max(@xs))[$x < max(@xs)] : $x
}

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Re^3: Finding the max()/min()
by Ven'Tatsu (Deacon) on Nov 11, 2004 at 22:02 UTC
    Your solution is calling max(@xs) twice each step. This leads to O(2^n) growth for what should be an O(n) problem. Modifying it slightly to call max once and save the value in a temp variable should save considerable time on long lists. 
    sub max {
    my ($x, @xs) = @_;
    @xs ? do { my $m = maxdo(@xs); ($x, $m)[$x < $m] } : $x;
}

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