zoukankan      html  css  js  c++  java
  • Longest Common Substring

    Problem Statement

    Give two string $s_1$ and $s_2$, find the longest common substring (LCS). E.g: X = [111001], Y = [11011], the longest common substring is [110] with length 3.

    One terse way is to use Dynamic Programming (DP) to analyze the complex problem.

    Instead of dealing with irregular substring, we can first deal with substring indexed by last character.

    Define $dp[i][j] =$ the length of longest common substring of $s_1[0$~$i]$ and $s_2[0$~$j]$ ending with $s1[i]$ and $s2[j]$.

    Then, the maximum LCS length could be the maximum number in array $dp$.

    In order to get the value of $dp[i][j]$, we need to know if $s1[i]$ == $s2[j]$. If it is, then the $dp[i][j] = dp[i-1][j-1]+1$, else it'll be zero. Thus:

    dp[i][j] = (s1[i] == s2[j] ? (dp[i-1][j-1] + 1) : 0);
    

    As we want to know the concrete string with LCM, we just need to do a few modifications.

    When we get a larger $dp[i][j]$ than present maxLength, we'll update the maxLength by $dp[i][j]$.

    if(dp[i][j] > maxLen)
        maxLen = dp[i][j];
    

    At the same time, we can also record the starting index of the new longer substring. For string $s_1$, the beginning index of LCM is the present index $i$ adding 1 minus the length of LCM, i.e.

    if(dp[i][j] > maxLen){
        maxLen = dp[i][j];
        maxIndex = i + 1 - maxLen; 
    }
    

    Finally, we need to initialize state of $dp$. That's simple:

    for(int i = 0; i < s1.length(); ++i)
        dp[i][0] = (s1[i] == s2[0] ? 1 : 0);
    
    for(int j = 0; j < s2.length(); ++j)
        dp[0][j] = (s1[0] == s2[j] ? 1 : 0);
    

    The complete code is:

    void LCM(const string s1, const string s2, int &sIndex, int &length)
    {
        n1 = s1.length();
        n2 = s2.length();
        
        if(0 == n1 || 0 == n2) 
        {
            sIndex = -1;
            length = 0;
            return;
        }
        
        // initialize dp
        vector<vector<int> > dp;
        for(int i = 0; i < n1; ++i){
            vector<int> tmp;
            tmp.push_back((s1[i] == s2[0] ? 1 : 0));  // Initialize the bottom line
            for(int j = 1; j < n2; ++j)
            {
                if(0 == i){
                    tmp.push_back((s1[0] == s2[j] ? 1 : 0));  // Initialize the left line
                }else{
                    tmp.push_back(0);  // Empty the interior area
                }
            }
            
            dp.push_back(tmp);
        }
        
        // compute max length and index
        length = 0;
        for(int i = 1; i < n1; ++i){
            for(int j = 1; j < n2; ++j){
                if(st1[i] == st2[j])
                    dp[i][j] = dp[i-1][j-1] + 1;
                    
                if(dp[i][j] > length){
                    length = dp[i][j];
                    sIndex = i + 1 - length;
                }
            }
        }    
    }
    
  • 相关阅读:
    给数组赋值nan
    loc和iloc的区别
    爬虫26-部署crawl爬虫
    爬虫25-scrapy框架详解
    爬虫24-scrapy框架部署
    爬虫23-验证码识别
    爬虫22-使用selenium爬取信息
    爬虫21-selenium用法
    爬虫20-浏览器自动运行简单方法
    爬虫19-线程生产者和消费者以及队列
  • 原文地址:https://www.cnblogs.com/kid551/p/4321392.html
Copyright © 2011-2022 走看看