Yeah, whatever. Here's a modification of the program.
Give the number of elements to extract as a command line
argument, and the elements to extract from on STDIN.
It assumes that there are enough elements on STDIN.
#!/usr/bin/perl
use strict;
use warnings;
my @heap;
my $M = @ARGV ? shift : 10;
sub heapify;
sub heapify {
my $idx = shift;
my $max = $idx;
for my $try (2 * $idx + 1, 2 * $idx + 2) {
$max = $try if $try < @heap && $heap [$try] > $heap [$max]
}
return if $max == $idx;
@heap [$idx, $max] = @heap [$max, $idx];
heapify $max;
}
sub extract_max {
return unless @heap;
my $min = $heap [0];
my $tmp = pop @heap;
if (@heap) {
$heap [0] = $tmp;
heapify 0;
}
return $min;
}
# First load the heap with initial data.
for (1 .. $M) {
push @heap => scalar <>;
}
chomp @heap;
#
# Heapify it.
#
for (my $i = int (@heap / 2); $i ;) {
heapify $i;
}
#
# Deal with the rest.
#
while (<>) {
chomp;
next if $_ >= $heap [0]; # It's too large.
$heap [0] = $_;
heapify 0;
}
my @result;
push @result => extract_max while @heap;
@result = reverse @result;
print "@result\n";
__END__
The running time of this is O (N log k), using O (k) memory.
Abigail
