Dav*_*d B 6 perl binary-search
我想在Perl中实现二进制搜索算法.我的'数组'按递减顺序排序(不是实际数组,而是获取索引并返回值的函数).问题是可能存在一系列相同的值.如果我的搜索值在这样的范围内,我想返回包含它的第一个索引.
这就是我写的:
# get_val should be a *decreasing* function for idexes $i in min..max,
# formally: for any $i,$j s.t. $max>=$i>$j>=$min :
# $get_val_subref($i, $extra) <= $get_val_subref($j, $extra)
# min and max are the inclusive boundaries for the search
# get_val sub should get an index in min..max and an extra data reference, and return
# the value for the given index
# returns the smallest index $i in min..max for which $get_val_subref($j, $extra)
# returns $searched_val, or undef if no such index exists
sub binary_search {
my ( $min, $max, $searched_val, $get_val_subref, $get_val_sub_extra_data )
= @_;
my ( $mid, $val );
while ( $min <= $max ) {
$mid = $min + int( ( $max - $min ) / 2 );
$val = $get_val_subref->( $mid, $get_val_sub_extra_data );
if ( $val > $searched_val ) {
$min = $mid + 1;
}
elsif ( $val < $searched_val ) {
$max = $mid - 1;
}
else { ## SEE MY QUESTION BELOW ##
# surely $val == $searched_val, but is it the first one?
if ( $mid > $min
and $get_val_subref->( $mid - 1, $get_val_sub_extra_data )
== $searched_val )
{
# $val == $searched_val and prev($val) == $searched_val
# we have to continue
$max = $mid - 1;
}
else {
# $val == $searched_val and prev($val) != $searched_val
# wer'e done
return $mid;
}
}
}
# $val was not found. return undef
return undef;
}
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这是一个使用它的简单示例:
sub get_val_sub {
my ( $pos, $a ) = @_;
my $val = $a->[$pos];
return $val;
}
my @arr = (80, 40, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0);
say "RET:", binary_search( 0, $#arr, 0, \&get_val_sub, \@arr );
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问题是我不确定我的最后一个(标有## SEE MY QUESTION BELOW ##)是"漂亮".有没有更好的方法呢?
虽然我最初同意 Axeman 的答案......它在某种程度上类似于我使用线性逻辑的第一个(非常糟糕的)答案(至少是其中的一小部分)。具体来说,没有理由用 来$get_val_subref调用$mid - 1。这是不必要的线性搜索步骤。
这就是我的建议。除了避免线性搜索之外,它还有一个非常简单的好处:
sub binary_search {
...
my ( $mid, $val, $solution );
while ( $min <= $max ) {
...
else {
$solution = $mid; # Store a possible solution.
$max = $mid - 1; # But continue with the binary search
# until $min and $max converge on each other.
}
}
return $solution;
}
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