我想编写一个简单的 perl 脚本来为给定的电话号码生成所有可能的单词。
我从数组的定义开始：
my @nums = (
['0'],
['1'],
['2', 'a', 'b', 'c'],
['3', 'd', 'e', 'f'],
['4', 'g', 'h', 'i'],
['5', 'j', 'k', 'l'],
['6', 'm', 'n', 'o'],
['7', 'p', 'q', 'r', 's'],
['8', 't', 'u', 'v'],
['9', 'w', 'x', 'y', 'z']
);
最终脚本应生成以下输出：
$ num2word 12
12
1a
1b
1c
$ num2word 213
213
21d
21e
21f
a13
a1d
a1e
a1f
b13
b1d
b1e
b1f
c13
c1d
c1e
c1f
我正在寻找可以完成大部分工作的任何模块（例如 List::Permutor 似乎不符合此任务的条件）。
有什么提示吗？谢谢！
我们自己的@brian d foy 用他的Set::CrossProduct模块解决了这个问题。
use Set::CrossProduct;
my $iterator = Set::CrossProduct->new(
[ [ qw(8 t u v) ], [ qw(0) ], [ qw(7 p q r s) ] ] );
print "@$_\n" for $iterator->combinations;
输出：
8 0 7
8 0 p
8 0 q
8 0 r
8 0 s
t 0 7
t 0 p
t 0 q
t 0 r
t 0 s
u 0 7
u 0 p
u 0 q
u 0 r
u 0 s
v 0 7
v 0 p
v 0 q
v 0 r
v 0 s
这可以满足您的要求。
use strict;
use warnings;
my @nums = (
[ qw/ 0 / ],
[ qw/ 1 / ],
[ qw /2 a b c / ],
[ qw /3 d e f / ],
[ qw /4 g h i / ],
[ qw /5 j k l / ],
[ qw /6 m n o / ],
[ qw /7 p q r s / ],
[ qw /8 t u v / ],
[ qw /9 w x y z / ],
);
list_matching('12');
list_matching('213');
sub list_matching {
my ($num) = @_;
my @num = $num =~ /\d/g;
my @map = (0) x @num;
do {
print join('', map { $nums[$num[$_]][$map[$_]] } 0 .. $#num), "\n";
my $i = $#map;
while ($i >= 0) {
last if ++$map[$i] < @{ $nums[$num[$i]] };
$map[$i--] = 0;
}
} while grep $_, @map;
}
输出
12
1a
1b
1c
213
21d
21e
21f
a13
a1d
a1e
a1f
b13
b1d
b1e
b1f
c13
c1d
c1e
c1f
实际上，确实有效，对我来说太早了......
use autodie;
use strict;
use warnings;
my @nums = (
['0'],
['1'],
['2', 'a', 'b', 'c'],
['3', 'd', 'e', 'f'],
['4', 'g', 'h', 'i'],
['5', 'j', 'k', 'l'],
['6', 'm', 'n', 'o'],
['7', 'p', 'q', 'r', 's'],
['8', 't', 'u', 'v'],
['9', 'w', 'x', 'y', 'z']
);
my $input = shift || die "Need a number!\n";
die "Input not numeric!\n" unless $input =~ m/^\d+$/;
my @digits = split //, $input;
my @rows;
push @rows, $nums[$_] for @digits;
print_row(0,'');
exit;
sub print_row {
my $i = shift;
my $word = shift;
my $row = $rows[$i];
for my $j (0..$#{$row}) {
my $word2 = $word . $row->[$j];
if ($i < $#rows) {
print_row($i+1, $word2);
}
else {
print "$word2\n";
}
}
}
无需模块：
my @nums = (
['0'],
['1'],
['2', 'a', 'b', 'c'],
['3', 'd', 'e', 'f'],
['4', 'g', 'h', 'i'],
['5', 'j', 'k', 'l'],
['6', 'm', 'n', 'o'],
['7', 'p', 'q', 'r', 's'],
['8', 't', 'u', 'v'],
['9', 'w', 'x', 'y', 'z']
);
print "$_\n" while glob join '', map sprintf('{%s}', join ',', @{$nums[$_]}), split //, $ARGV[0]
use strict;
use warnings;
my @nums = (
['0'], ['1'], ['2', 'a', 'b', 'c'],
['3', 'd', 'e', 'f'], ['4', 'g', 'h', 'i'],
['5', 'j', 'k', 'l'], ['6', 'm', 'n', 'o'],
['7', 'p', 'q', 'r', 's'], ['8', 't', 'u', 'v'],
['9', 'w', 'x', 'y', 'z']);
num2word(12);
num2word(213);
sub num2word {
my ($i, $n, $t) = ($_[0]=~/(.)(.*)/, $_[1]);
for (@{$nums[$i]}) {
print "$t$_\n" and next if !length($n);
num2word($n, defined $t ? $t.$_ : $_);
}
}
一个基本的递归解决方案：
#!/usr/bin/perl
use strict;
use warnings;
my $phone_number = $ARGV[0] or die "No phone number";
my @nums = (
['0'],
['1'],
[ '2', 'a', 'b', 'c' ],
[ '3', 'd', 'e', 'f' ],
[ '4', 'g', 'h', 'i' ],
[ '5', 'j', 'k', 'l' ],
[ '6', 'm', 'n', 'o' ],
[ '7', 'p', 'q', 'r', 's' ],
[ '8', 't', 'u', 'v' ],
[ '9', 'w', 'x', 'y', 'z' ]
);
my %letters = map { shift @{$_} => $_ } @nums;
my @permutations;
sub recurse {
my $str = shift;
my $done = shift || '';
unless ($str) {
push @permutations, $done;
return;
}
my $next = substr( $str, 0, 1 );
$str = substr( $str, 1 );
recurse( $str, $done . $next );
if ( my @chars = @{ $letters{$next} } ) {
recurse( $str, $done . $_ ) foreach @chars;
}
}
recurse($phone_number);
print "$_\n" foreach @permutations;
和：
perl num2word 12
12
1a
1b
1c
perl num2word 213
213
21d
21e
21f
a13
a1d
a1e
a1f
b13
b1d
b1e
b1f
c13
c1d
c1e
c1f