Question

我有一个字符串，用于描述像这样的n x m个元素的矩阵：

§inputmap = "
~~~~~~~~~~~~~~~~~~~~B~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~BBB........BBB~~~~~~~~~~~~~
~~~~~~~~~~BB...............FBB~~~~~~~~~~
~~~~~~~~BB....................BB~~~~~~~~
~~~~~~BB.....F..................BB~~~~~~
~~~~~BB.....................F.....B~~~~~
~~~~B..............................B~~~~
~~~B........F.......................B~~~
~~BB.........F......................BB~~
~~B................F.................BB~
~BF....F....F........................FB~
~B.....................................B
B.....................................FB
B........F......F......................B
B...........................F..........B
B......................................B
B......................................B
B.......F.......................F......B
B......FFF.............................B
B.......F.............................FB
~B..................F.................FB
~BF...........................F.......B~
~~B...F...........F..........FFFFF.F.BB~
~~BB..................F..F....F.....BB~~
~~~B.......................FF.FF....B~~~
~~~~B..............................B~~~~
~~~~~BB...........................B~~~~~
~~~~~~BB........................BB~~~~~~
~~~~~~~~BB..........F..........B~~~~~~~~
~~~~~~~~~~BB................BB~~~~~~~~~~
~~~~~~~~~~~~~BBB.......F.BBB~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~BBBBBB~~~~~~~~~~~~~~~~~
";
$inputmap = trim($inputmap);

我需要构建一个正则表达式（或其他东西）来搜索字符串：

$search = "
*F*
FFF
*F*
";
$search = trim($search);

整个网格。在另一方面，我需要找到一个5个不同字母“F”（3个垂直和3个水平）的模式，返回地图上找到的模式的行/列位置。

考虑到输入矩阵可能不同（5x5或10x10或20x25或......），有没有办法用php和正则表达式解决我的问题？

Answer 1

您可以使用$length=strstr($inputmap,"\n")查找每行的宽度。然后，您可以构建一个正则表达式，该表达式将找到F，后跟($length-2)个其他字符，后跟3个F，后跟($length-2)个其他字符，后跟F。

Answer 2

你可以这样做（没有正则表达式）：

$map = explode("\n", $inputmap);

for ($vcount = 0; $vcount < sizeof($map); ++$vcount) {  // Loop through the map vertically
    for ($hcount = 0; $hcount < strlen($map[$vcount]); ++$hcount) { // Loop through each character of each line
        if ($map[$vcount][$hcount] == "F") {
            if ($map[$vcount + 1][$hcount - 1] == "F" && $map[$vcount + 1][$hcount] == "F" && 
                $map[$vcount + 1][$hcount + 1] == "F" && $map[$vcount + 2][$hcount] == "F")
                echo "Pattern found, starting at : (v)$vcount x (h)$hcount";
        }
    }
}

$> php test.php
php test.php
Pattern found, starting at : (v)18 x (h)8
Pattern found, starting at : (v)22 x (h)30
$>

但我必须承认，超大地图可能需要一段时间。

/！\添加一些代码以验证行$vcount + 1/2实际存在，以及char $hcount + 1。

Answer 3

您可以使用以下代码：

$lines = array_filter(preg_split("#\r\n?|\n#", $string)); // Creating array of lines
$matrix = array_map('str_split', $lines); // Creating a matrix

foreach($lines as $line_number => $line_content){ // Looping through the lines
    $pos = strpos($line_content, 'FFF');
    if(!$pos === false){// If FFF found
        while(true){
            if(isset($matrix[$line_number-1][$pos+1],$matrix[$line_number+1][$pos+1]) && $matrix[$line_number-1][$pos+1] == 'F' && $matrix[$line_number+1][$pos+1] == 'F'){ //Checking ...
                echo 'Found at: X:'.$pos.' & Y:'.$line_number.'<br>'; // Ouput
            }
            $pos = strpos($line_content, 'FFF', $pos+1); // Search further
            if(!is_int($pos)){
                break;
            }
        }
    }
}

发生了什么事？

我们创建一个行数组
我们创建一个矩阵
我们遍历行并搜索FFF这是为了提高性能。因此，我们不是循环遍历整个矩阵并搜索F，而是直接搜索FFF
检查矩阵中是否存在该值（以防止未定义的索引错误），然后检查它是否等于F
输出
在同一行上进一步搜索

<强> Online demo

_{请注意，坐标是+}的中心

Answer 4

尝试这种模式：

如果行尾为\n：

$pattern = '~F.{'. ($n-2) . '}FFF.{' . ($n-2) . 'F~s';

如果行尾为\r\n，则替换为$n-3

或者更好地使用quinxorin技巧来知道\n

之前的行长度

Answer 5

@FrankieTheKneeMan＆amp;所有......我最初的方法是为输入网格图的'n'字符构建'n'正则表达式。有点像...

$search_001 = "
.F......................................
FFF.....................................
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";


$search_002 = "
..F.....................................
.FFF....................................
..F.....................................
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";

（... omissis ...）

$search_100 = "
........................................
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..........F.............................
.........FFF............................
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";

（... omissis ...）

$search_999 = "
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......................................F.
.....................................FFF
......................................F.
";

其中“。” obvioulsy意味着任何角色。

这是一个愚蠢的想法吗？！？

php垂直正则表达式搜索

5 个答案: