我有一个函数,binary_range_search
,它是这样调用的:
my $brs_iterator = binary_range_search(
target => $range, # eg. [1, 200]
search => $ranges # eg. [ {start => 1, end => 1000},
); # {start => 500, end => 1500} ]
brs_iterator->()
将遍历 $range 重叠的所有 @$ranges。
我想扩展 binary_range_search
以便能够以多个范围作为目标来调用它,例如:
target => $target_ranges # eg. [ [1, 200], [50, 300], ... ]
search => $search_ranges # as above
因此,当对 $range->[0] 的搜索用完时,它应该继续搜索 $range->[1],依此类推。这是有问题的函数,其原始形式:
sub binary_range_search {
my %options = @_;
my $range = $options{target} || return;
my $ranges = $options{search} || return;
my ( $low, $high ) = ( 0, @{$ranges} - 1 );
while ( $low <= $high ) {
my $try = int( ( $low + $high ) / 2 );
$low = $try + 1, next if $ranges->[$try]{end} < $range->[0];
$high = $try - 1, next if $ranges->[$try]{start} > $range->[1];
my ( $down, $up ) = ($try) x 2;
my %seen = ();
my $brs_iterator = sub {
if ( $ranges->[ $up + 1 ]{end} >= $range->[0]
and $ranges->[ $up + 1 ]{start} <= $range->[1]
and !exists $seen{ $up + 1 } )
{
$seen{ $up + 1 } = undef;
return $ranges->[ ++$up ];
}
elsif ( $ranges->[ $down - 1 ]{end} >= $range->[0]
and $ranges->[ $down + 1 ]{start} <= $range->[1]
and !exists $seen{ $down - 1 }
and $down > 0 )
{
$seen{ $down - 1 } = undef;
return $ranges->[ --$down ];
}
elsif ( !exists $seen{$try} ) {
$seen{$try} = undef;
return $ranges->[$try];
}
else {
return;
}
};
return $brs_iterator;
}
return sub { };
}
这是一个标准的二分搜索策略,直到它找到一个重叠的范围。然后它向右移动,耗尽它,向左移动,耗尽它,最后放弃。理想情况下,我想它应该 shift
下一个目标范围,然后重新搜索(也许通过递归?)。我的问题是,我不确定如何使用迭代器构造使其工作。
最佳答案
我只是将您的迭代器生成包装在一个 for 循环中,并构建了一个迭代器函数数组。
根据上下文,我要么返回一个主迭代器,要么返回一个迭代器函数列表。我不确定你想要什么。
use strict;
use warnings;
my $t = [ [1,200], [400,900] ];
my @r = (
{ start => 1, end => 100 },
{ start => 2, end => 500 },
{ start => 204, end => 500 },
{ start => 208, end => 500 },
{ start => 215, end => 1000 },
{ start => 150, end => 1000 },
{ start => 500, end => 1100 },
);
# Get a master iterator that will process each iterator in turn.
my $brs_iterator = binary_range_search(
targets => $t,
search => \@r,
);
# Get an array of iterators
my @brs_iterator = binary_range_search(
targets => $t,
search => \@r,
);
sub binary_range_search {
my %options = @_;
my $targets = $options{targets} || return;
my $ranges = $options{search} || return;
my @iterators;
TARGET:
for my $target ( @$targets ) {
my ( $low, $high ) = ( 0, $#{$ranges} );
RANGE_CHECK:
while ( $low <= $high ) {
my $try = int( ( $low + $high ) / 2 );
# Remove non-overlapping ranges
$low = $try + 1, next RANGE_CHECK
if $ranges->[$try]{end} < $target->[0];
$high = $try - 1, next RANGE_CHECK
if $ranges->[$try]{start} > $target->[1];
my ( $down, $up ) = ($try) x 2;
my %seen = ();
my $brs_iterator = sub {
if ( exists $ranges->[$up + 1]
and $ranges->[ $up + 1 ]{end} >= $target->[0]
and $ranges->[ $up + 1 ]{start} <= $target->[1]
and !exists $seen{ $up + 1 } )
{
$seen{ $up + 1 } = undef;
return $ranges->[ ++$up ];
}
elsif ( $ranges->[ $down - 1 ]{end} >= $target->[0]
and $ranges->[ $down + 1 ]{start} <= $target->[1]
and !exists $seen{ $down - 1 }
and $down > 0 )
{
$seen{ $down - 1 } = undef;
return $ranges->[ --$down ];
}
elsif ( !exists $seen{$try} ) {
$seen{$try} = undef;
return $ranges->[$try];
}
else {
return;
}
};
push @iterators, $brs_iterator;
next TARGET;
}
}
# In scalar context return master iterator that iterates over the list of range iterators.
# In list context returns a list of range iterators.
return wantarray
? @iterators
: sub {
while( @iterators ) {
if( my $range = $iterators[0]() ) {
return $range;
}
shift @iterators;
}
return;
};
}
关于perl - 如何扩展二进制搜索迭代器以使用多个目标,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2046390/