LeetCode Regular Expression Matching 网上一个不错的实现（非递归）

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

以下是牛人的实现：

class Solution {
public:
    bool isMatch(string s, string p) {
        int i, j;
        int m = s.size();
        int n = p.size();
        const char *s1 = s.c_str();
        const char *p1 = p.c_str();
        /**
         * b[i + 1][j + 1]: if s[0..i] matches p[0..j]
         * if p[j] != '*'
         * b[i + 1][j + 1] = b[i][j] && s[i] == p[j]
         * if p[j] == '*', denote p[j - 1] with x,
         * then b[i + 1][j + 1] is true iff any of the following is true
         * 1) "x*" repeats 0 time and matches empty: b[i + 1][j -1]
         * 2) "x*" repeats 1 time and matches x: b[i + 1][j]
         * 3) "x*" repeats >= 2 times and matches "x*x": s[i] == x && b[i][j + 1]
         * '.' matches any single character
         */
        bool b[m + 1][n + 1];
        b[0][0] = true;
        for (i = 0; i < m; i++) 
        {
            b[i + 1][0] = false;
        }
        // p[0..j - 2, j - 1, j] matches empty iff p[j] is '*' and p[0..j - 2] matches empty
        for (j = 0; j < n; j++) 
        {
            b[0][j + 1] = j > 0 && '*' == p1[j] && b[0][j - 1];
        }

        for (i = 0; i < m; i++) 
        {
            for (j = 0; j < n; j++) 
            {
                if (p[j] != '*')
                {
                    b[i + 1][j + 1] = b[i][j] && ('.' == p1[j] || s1[i] == p1[j]);
                } 
                else 
                {
                    b[i + 1][j + 1] = b[i + 1][j - 1] && j > 0 || b[i + 1][j] ||
                                b[i][j + 1] && j > 0 && ('.' == p1[j - 1] || s1[i] == p1[j - 1]);
                }
            }
        }
        return b[m][n];
    }


};

这应该是用动态规划的方法实现的。