[Leetcode]Dynamic Programming

Posted 2022-12-07 stary_yan

[Leetcode]Dynamic Programming

Regular Expression Matching

Analysis

It is easy to understand the problem. What we need to do is to find if the string can be matched by the pattern. At first, I think it will be easy to use FSM (finite state machine) just like the compilers do. But I soon find that it will be a bit hard to implement because we have to define the transition table. However, we can use dynamic programming to solve the problem instead of FSM.

We can define dp[i][j] as the state which represents if s[0:j] can be matched by patten[0:i] or not.

dp[i][j] =
    if pattern[i] != '*':
        if pattern[i] == s[j] || pattern[i] == '.':
            dp[i - 1][j - 1]
        else:
            false
    if pattern[i] == '*' :
         if pattern[i - 1] == '.': # Match any string
            dp[i - 2][0] || dp[i - 2][1] || ..... || dp[i - 2][j]
         else:
            dp[i - 2][j] || dp[i - 2][j - 1] (if pattern[i - 1] == s[j - 1]

Code

#include <iostream>
#include <map>
#include <vector>
using namespace std;

class Solution 
public:
    bool isMatch(string s, string p) 
        vector<vector<bool>> dp(p.size() + 1, vector<bool>(s.size() + 1, false));
        dp[0][0] = true;
        for (int i = 1; i <= p.size(); i++) 
            if (p[i - 1] == '*')  dp[i][0] = dp[i - 2][0]; 
        
        for (int i = 1; i <= p.size(); i++) 
            for (int j = 1; j <= s.size(); j++) 
                if (p[i - 1] != '*') 
                    if (p[i - 1] == '.' || p[i - 1] == s[j - 1]) 
                        dp[i][j] = dp[i - 1][j - 1];
                     else 
                        dp[i][j] = false;
                    
                 else 
                    if (p[i - 2] == '.') 
                        int m = j;
                        while (m >= 0 && dp[i][j] == false) 
                            dp[i][j] = dp[i - 2][m];
                            m -= 1;
                        
                     else 
                        dp[i][j] = dp[i - 2][j];
                        int m = j;
                        while (m > 0 && dp[i][j] == false) 
                            if (s[m - 1] == p[i - 2]) 
                                dp[i][j] = dp[i - 2][m - 1];
                                m -= 1;
                             else 
                                break;
                            
                        
                    
                
            
        
        return dp[p.size()][s.size()];
    
;

int main() 
    Solution so;
    cout << so.isMatch("baabbbaccbccacacc", "c*..b*a*a.*a..*c") << endl;
    return 0;

There are some tricks which you will have to consider by yourself.

Time complexity: O(n * m)