Question

我有一个字符串（比如字符串1）需要与其他字符串匹配（string2）两个字符串都具有相同的长度，并且不区分大小写。

我想打印两个字符串之间的字符匹配数。

Eg: String 1: stranger
    String 2: strangem

     Match count = 7

我试过了：

$string1 = "stranger";
$string2 = "strangem";
my $count = $string1 =~ m/string2/ig;

print "$count\n";

感谢任何形式的帮助。在此先感谢!!

Answer 1

排他或，然后计算零字符（字符串相同）：

my $string1 = "stranger";
my $string2 = "strangem";
my $count = ( lc $string1 ^ lc $string2 ) =~ tr/\0//;

print "$count\n";

更新：错过了“案例不敏感”位。

Answer 2

您可以使用substr：

#!/usr/bin/perl
use warnings;
use strict;

my $string1=lc('stranger');
my $string2=lc('strangem');
my $count=0;
for (0..length($string1)-1) {
    $count++ if substr($string1,$_,1) eq substr($string2,$_,1);
}
print $count; #prints 7

或者您可以使用split将所有字符作为数组获取，并循环：

#!/usr/bin/perl
use warnings;
use strict;

my $string1=lc('stranger');
my $string2=lc('strangem');
my $count=0;
my @chars1=split//,$string1; 
my @chars2=split//,$string2;
for (0..$#chars1) {
    $count++ if $chars1[$_] eq $chars2[$_];
}
print $count; #prints 7

（fc提供比lc更准确的结果，但我追求向后兼容。）

Answer 3

未经测试

sub cm
{
    my @a = shift;
    my @b = shift;

    # First match prefix of string:
    my $n = 0;
    while ($n < $#a && $n < $#b && $a[$n] eq $b[$n]) {
       ++$n;
    }
    # Then skip one char on either side, and recurse.
    if ($n < $#a && $n < $#b) {
       # Match rest by skipping one place:
       my $n2best = 0;
       my $n2a = cm(splice(@a, $n), splice(@b, $n + 1));
       $n2best = $n2a;
       my $n2b = cm(splice(@a, $n + 1), splice(@b, $n));
       $n2best = $n2b if $n2b > $n2best;
       my $n2c = cm(splice(@a, $n + 1), splice(@b, $n + 1));
       $n2best = $n2c if $n2c > $n2best;
       $n += $n2best;
    }
    return $n;
}

sub count_matches
{
    my $a = shift;
    my $b = shift;

    my @a_chars = split //, $a;
    my @b_chars = split //, $b;
    return cm(@a_chars, @b_chars); 
}

print count_matches('stranger', 'strangem')

perl中字符串之间的字符匹配计数

3 个答案: