是否有用于在陣列中查找重復子陣列（確切順序）的內置Perl函式？

假設陣列是 (1,2,3,4,5,6,7,8,9)，另一個子陣列是 (2,3,4) 有沒有檢查子陣列模式是否的函式（完全精確的順序）存在于陣列中？在這種情況下，它將回傳任何表明它存在的指標（索引）。如果陣列中存在多個子陣列，例如 (4,2,3,4,2,3,4)，也需要處理重復項。如果它碰巧匹配多次，例如： Array = (2,3,2,3,2,2,3,2) Sub Array = (2,3,2) 將按順序回傳匹配的起始索引：0 ,2,5 或者如果它洗掉，將導致 (3,2)

編輯：元素不必是 num

uj5u.com熱心網友回復：

沒有內置方法，但很容易撰寫：

#!/usr/bin/env perl
use warnings;
use strict;
use feature qw/say/;

# Takes two arrayrefs of numbers.
#
# Returns the first index in the first one where the second list appears, or
# -1 if not found.
sub find_sublist(  ) {
  my ($haystack, $needle) = @_;
  my $nlen = @$needle;
  my $hlen = @$haystack;
  return -1 if $hlen == 0 || $nlen == 0;
 HAYSTACK_POS:
  for (my $n = 0; $n <= $hlen - $nlen; $n  ) {
    for (my $m = 0; $m < $nlen; $m  ) {
      if ($haystack->[$n   $m] != $needle->[$m]) {
        next HAYSTACK_POS;
      }
    }
    return $n;
  }
  return -1;
}


# Takes two arrayrefs of numbers.
#
# Returns a list of the starting indexes of the first list
# of every run of the second list. Returns an empty list if
# there are no matches.
sub find_sublists(  ) {
  my ($haystack, $needle) = @_;
  my $nlen = @$needle;
  my $hlen = @$haystack;
  my @positions;
  return @positions if $hlen == 0 || $nlen == 0;
 HAYSTACK_POS:
  for (my $n = 0; $n <= $hlen - $nlen; $n  ) {
    for (my $m = 0; $m < $nlen; $m  ) {
      if ($haystack->[$n   $m] != $needle->[$m]) {
        next HAYSTACK_POS;
      }
    }
    push @positions, $n;
  }
  return @positions;
}

# Takes two arrayrefs of numbers.
#
# Returns a new list that is the first one with every non-overlapping run of
# the second second list removed.
sub remove_sublists(  ) {
  my @haystack = @{$_[0]};
  my $needle = $_[1];
  while ((my $pos = find_sublist @haystack, $needle) != -1) {
    splice @haystack, $pos, @$needle;
  }
  return @haystack;
}

my @list1 = (1,2,3,4,5,6,7,8,9);
my @list2 = (4,2,3,4,2,3,4);
my @list3 = (2,3,2,3,2,2,3,2);
say find_sublist(@list1, [2, 3, 4]);            # Returns 1
say find_sublist([2,9,3,4], [2,3,4]);           # Returns -1
my @positions = find_sublists(@list2, [2,3,4]); # 1,4
say join(",", @positions);
@positions = find_sublists(@list3, [2,3,2]); # 0,2,5
say join(",", @positions);
say join(",", remove_sublists(@list1, [2,3,4])); # 1,5,6,7,8,9
say join(",", remove_sublists(@list3, [2,3,2])); # 3,2

uj5u.com熱心網友回復：

如果輸入是由您perl的整數表示的數字（如圖所示），您可以使用

# Indexes
my $pattern = pack "W*", @pattern;
my $array   = pack "W*", @array;
my @indexes;
push @indexes, $-[0] while $array =~ /\Q$pattern/g;

# Removal
my $pattern = pack "W*", @pattern;
my $array   = pack "W*", @array;
$array =~ s/\Q$pattern//g;
@array = unpack "W*", $array;

它如何處理重疊。

           /---\     /---\   Removed
2,3,2 from 2,3,2,3,2,2,3,2 
               \---/         Not removed

請注意，如果您可以將輸入映射到數字，這也有效。

my ( %map_f, @map_r );
for ( @array, @pattern ) {
   if ( !exists{ $map{ $_ } } ) {
      $map_f{ $_ } = @map_r;
      push @map_r, $_;
   }
}

my $pattern = pack "W*", @map_f{ @pattern };
my $array   = pack "W*", @map_f{ @array   };
$array =~ s/\Q$pattern//g;
@array = @map_r[ unpack "W*", $array ];

這不是最好的演算法，但通過將作業從 Perl 轉移到正則運算式引擎，它應該非常快。

標籤：数组 perl

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