PHP algorithm: PHP implements the problem of the longest common substring

What this article brings to you is the problem of the longest common substring in PHP implementation of the PHP algorithm. It has certain reference value. Friends in need can refer to it. I hope it will be helpful to you.

Longest common substring problem:

Given two strings, find the length of the longest identical substring between them.

Violent solution ideas:

1. Start with each character of the two strings and compare them later, which will require two levels of loops

2. The comparison method inside the two-level loop is also a one-level loop, starting from the current character, traversing and comparing until there is a difference, then jump out of the loop and record the length of the same substring

3. Based on the longest length, there are three levels of loops. Time complexity O(n^3)

longest=0
for i=0;i<str1.size;i++
    for j=0;j<str2.size;j++
        m=i  n=j length=0
    while(m<str1.size && n<str2.size)
            if str1[m]!=str2[n] break
            ++length
            ++m
            ++n
        longest=longest<length ? length:longest

Dynamic programming method:

1. In the above comparison process, starting from i and j, if After reaching a different stop, the next starting position will be compared repeatedly

2. Dynamic programming method-space for time, matrix graph, can reduce the complexity to O(n^2)

3.str1 is the horizontal axis, str2 is the vertical axis, table[i][j] is the length of the longest substring ending with these two characters

4.table[0][j] It can be deduced, if str1[0]==str2[j] is 1, table[i][0] can be deduced, if str1[i]==str2[0] is 1, and the rest is 0

5.table[i][j] If str1[i]==str2[j] can be obtained from table[i-1][j-1] 1, if not equal to 0

Assume that the two strings are s and t respectively, s[i] and t[j] represent the i-th and j-th characters respectively (the character order starts from 0), and then let L[i, j] represent s[ i] and s[j] are the maximum length of the same substring at the end. It should not be difficult to derive the relationship between L[i, j] and L[i 1,j 1], because the only difference between the two is the pair of characters s[i 1] and t[j 1]. If s[i 1] and t[j 1] are different, then L[i 1, j 1] should naturally be 0, because any substring ending with them cannot be exactly the same; and if s[i 1] The same as t[j 1], then just add these two characters after the longest identical substring ending with s[i] and t[j], so that the length can be increased by one. Combining the above two situations, we get the relationship L[i 1,j 1]=(s[i]==t[j]?L[i,j] 1:0).

Code example:

<?php
$str1="abcdef";
$str2="esdfdbcde1";

//暴力解法
function longestCommonSubstring1($str1,$str2){
        $longest=0;
        $size1=strlen($str1);
        $size2=strlen($str2);
        for($i=0;$i<$size1;$i++){
                for($j=0;$j<$size2;$j++){
                        $m=$i;
                        $n=$j;
                        $length=0;
                        while($m<$size1 && $n<$size2){
                                if($str1[$m]!=$str2[$n]) break;
                                ++$length;
                                ++$m;
                                ++$n;
                        }   
                        $longest=$longest < $length ? $length : $longest;
                }   
        }   
        return $longest;
}
//矩阵动态规划法
function longestCommonSubstring2($str1,$str2){
        $size1=strlen($str1);
        $size2=strlen($str2);
        $table=array();
        for($i=0;$i<$size1;$i++){
                $table[$i][0]=$str1[$i]==$str2[0] ? 1:0;
        }   
        for($j=0;$j<$size2;$j++){
                $table[0][$j]=$str1[0]==$str2[$j] ? 1:0;
        }   
        for($i=1;$i<$size1;$i++){
                for($j=1;$j<$size2;$j++){
                        if($str1[$i]==$str2[$j]){
                                $table[$i][$j]=$table[$i-1][$j-1]+1;    
                        }else{
                                $table[$i][$j]=0;
                        }    
                }   
        }   
        $longest=0;
        for($i=0;$i<$size1;$i++){
                for($j=0;$j<$size2;$j++){
                        $longest=$longest<$table[$i][$j] ? $table[$i][$j] : $longest;
        }}  
        return $longest;
}
$len=longestCommonSubstring1($str1,$str2);
$len=longestCommonSubstring2($str1,$str2);
var_dump($len);

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