[LeetCode]Regular Expression Matching

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本文介绍了两种实现正则表达式匹配的算法：递归方法和动态规划方法。递归方法通过判断字符是否匹配及特殊符号 '*' 的使用来进行匹配；动态规划方法则利用二维数组记录中间结果避免重复计算，提高效率。

class Solution {
//1. If the next character of p is NOT '*', then it must match the current character of s. 
//Continue pattern matching with the next character of both s and p.
//2. If the next character of p is '*', then we do a brute force exhaustive matching of 0, 1, or 
//more repeats of current character of p... Until we could not match any more characters.
//recursion
public:
	bool isMatch(const char *s, const char *p) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function    
		if (*p == '\0') return *s == '\0';
		if ( *(p+1) != '*')
			return ( (*s == *p) || (*p == '.' && *s != '\0') ) && isMatch(s+1, p+1);

		//if *(p+1) is '*', and *s == *p
		while ( (*s == *p) || (*p == '.' && *s != '\0') )
		{
			if( isMatch(s, p+2) ) return true;//pass zero 
			s++;//pass as more as possible
		}
		return isMatch(s, p+2);
	}
};

second time

class Solution {
//tabulation DP based
public:
    bool isMatch(const char *s, const char *p) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        int len1 = 0;
        for(const char* q = s; *q != '\0' && q != NULL; ++q) len1++;
        int len2 = 0;
        for(const char* q = p; *q != '\0' && q != NULL; ++q) len2++;
        
        vector<vector<bool> > f(len1+1, vector<bool>(len2+1, false));
        f[0][0] = true;
        for(int i = 0; i <= len1; ++i)
        {
            char preChar = '\0';
            int preIdx = 0;
            for(int j = 1; j <= len2; ++j)
            {
                if(i >= 1 && (p[j-1] == '.' || s[i-1] == p[j-1])) f[i][j] = f[i-1][j-1];
                else if(p[j-1] == '*')
                {
                    if(i >= 1 && (preChar == s[i-1] || preChar == '.')) f[i][j] = f[i-1][j] || f[i][j-1];
                    f[i][j] = f[i][j] || f[i][preIdx];
                    //f[i-1][j] = f[i-1][j] || f[i-1][preIdx];
                }
                if(p[j-1] != '*') preChar = p[j-1], preIdx = j-1;
            }
        }        
        
        return f[len1][len2];
    }
};