如何计算Perl中的重叠子字符串？ [英] How can I count overlapping substrings in Perl?

问题描述

我需要实现一个程序来计数在perl中的字符串中子串的出现。我已经实现如下

i need to implement a program to count the occurrence of a substring in a string in perl. i have implemented it as follows

sub countnmstr
{
  $count =0;
  $count++ while $_[0] =~ /$_[1]/g;
  return $count;
}

$count = countnmstr("aaa","aa");

print "$count\n";

现在这是我通常会做的。然而，在上面的实现我想计算'aa'在'aaa'的发生。这里我得到答案为1这似乎合理，但我需要考虑重叠的情况下，以及。因此，上述情况应该给出答案为2，因为如果我们考虑重叠，有两个aa。

now this is what i would normally do. however, in the implementation above i want to count occurrence of 'aa' in 'aaa'. here i get answer as 1 which seems reasonable but i need to consider the overlapping cases as well. hence the above case should give an answer as 2 since there are two 'aa's if we consider overlap.

任何人都可以建议如何实现这样的功能。

can anyone suggest how to implement such a function??

推荐答案

请参阅 ysth的回答 ...我没有意识到该模式可以仅由零宽度断言组成，仍然可以用于此目的。

See ysth's answer ... I failed to realize that the pattern could consist solely of a zero width assertion and still work for this purpose.

您可以使用 positive lookahead ，并将函数写为：

You can use positive lookahead as suggested by others, and write the function as:

sub countnmstr {
    my ($haystack, $needle) = @_;
    my ($first, $rest) = $needle =~ /^(.)(.*)$/;
    return scalar (() = $haystack =~ /(\Q$first\E(?=\Q$rest\E))/g);
}

您也可以使用 pos ，以调整下一次搜索的起点：

You can also use pos to adjust where the next search picks up from:

#!/usr/bin/perl

use strict; use warnings;

sub countnmstr {
    my ($haystack, $needle) = @_;
    my $adj = length($needle) - 1;
    die "Search string cannot be empty!" if $adj < 0;

    my $count = 0;
    while ( $haystack =~ /\Q$needle/g ) {
        pos $haystack -= $adj;
        $count += 1;
    }
    return $count;
}

print countnmstr("aaa","aa"), "\n";

输出：

C:\Temp> t
2

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