Sword finger offer19 regular expression matching string dynamic programming

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Note：

f[i][j] Express s front i Characters and p front j Whether characters match
For convenience , Add a space before each string , Subscript becomes from 1 Start ,1 It means there is a match 1 Characters

Then it is to see whether it is according to the situation * Discussion by situation
No * It's easy to handle , Just write it directly , But as for * You have to enumerate
Look at it separately * Express 0 Characters 1 Characters …

Optimize this O(n) Complexity , Yes, it is f[i - 1][j] A comparison will show that he happens to be f[i][j] And s[i] & s[j-1] Operate

The code is as follows ：

class Solution {
    
public:
    bool isMatch(string s, string p) {
    
        int n = s.size(), m = p.size();
        s = ' ' + s, p = ' ' + p;
        vector<vector<bool>> f(n + 1, vector<bool>(m + 1));
        f[0][0] = true;

        for(int i = 0; i <= n; i ++)
            for(int j = 1; j <= m; j ++){
    
                if(j + 1 <= m && p[j + 1] == '*') continue;
                if(i && p[j] != '*') 
                    f[i][j] = f[i - 1][j - 1] && (s[i] == p[j] || p[j] == '.');
                else if(p[j] == '*'){
    
                    f[i][j] = f[i][j - 2] || i && f[i - 1][j] && (s[i] == p[j - 1] || p[j - 1] == '.');
                }
            }
        return f[n][m];
    }
};