最短编辑距离
问题：
由字符串 s1 通过下列三种操作
1、插入一个字符；
2、删除一个字符；
3、改变一个字符
变换到字符串 s2 所需要的最少操作次数（亦即最短编辑距离问题）
php代码实现如下：
<?php
class levenshtein_distance
{
private $source;
private $target;
public function __construct($source, $target)
{
$this->source = $source;
$this->target = $target;
}
public function get_ld()
{
$source = $this->source;
$target = $this->target;
$cell = mb_strlen($source, 'gbk');
$row = mb_strlen($target, 'gbk');
if ($cell == 0)
{
return $row;
}
if ($row == 0)
{
return $cell;
}
$matrix = array();
for ($i = 0; $i <= $row; $i++)
{
for ($j = 0; $j <= $cell; $j++)
{
if($i == 0)
{
$matrix[0][$j] = $j;
}
else if($i > 0 && $j == 0)
{
$matrix[$i][0] = $i;
}
else
{
$matrix[$i][$j] = 0;
}
}
}
for ($m = 0; $m < $row; $m++)
{
for ($n = 0; $n < $cell; $n++)
{
if ($source[$n] == $target[$m])
{
$tmp = 0;
}
else
{
$tmp = 1;
}
$matrix[$m + 1][$n + 1] = min($matrix[$m][$n] + $tmp, $matrix[$m + 1][$n] + 1, $matrix[$m][$n + 1] + 1);
}
}
return $matrix[$row][$cell];
}
}
if(!function_exists('mb_strlen'))
{
function mb_strlen($str, $charset)
{
$charsets["utf-8"] = $charsets["utf8"] = "/[\x01-\x7f]|[\xc2-\xdf][\x80-\xbf]|[\xe0-\xef][\x80-\xbf]{2}|[\xf0-\xff][\x80-\xbf]{3}/";
$charsets["gb2312"] = "/[\x01-\x7f]|[\xb0-\xf7][\xa0-\xfe]/";
$charsets["gbk"] = "/[\x01-\x7f]|[\x81-\xfe][\x40-\xfe]/";
$charsets["big5"] = "/[\x01-\x7f]|[\x81-\xfe]([\x40-\x7e]|\xa1-\xfe])/";
preg_match_all($charsets[$charset], $str, $matches); #按照指定编码将字符串切分成一个数组
return count($matches[0]);
}
}
function read()
{
fscanf(STDIN, "%s\n", $input);
$input = trim($input);
return $input;
}
function test()
{
$source = 'abc';
$target = 'ab';
$ld = new levenshtein_distance($source, $target);
echo '以上两个字符串的最短编辑距离为：'.$ld->get_ld();
}
function main()
{
echo iconv('utf-8', 'gbk', "请输入源字符串\n");
$source = read();
echo iconv('utf-8', 'gbk', "请输入目标字符串\n");
$target = read();
$ld = new levenshtein_distance($source, $target);
echo iconv('utf-8', 'gbk', '以上两个字符串的最短编辑距离为：'.$ld->get_ld()."\n");
}
//test();
main();