Regular Expression Matching

题目描述：

Implement regular expression matching with support for '.' and '*'.

---

'.' 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

---

这道题难度等级：Hard。我看了提示，说要使用Dynamic Programming、Backtracking。恶补了一下这两个方面的知识，结果还是没做出来。后来想到了递归，大体思路已经正确，可是细节之处总是处理不好。此题，寡人败了。
我查阅了很多文章，下面的解法在简洁性和易懂性上，无人能出其右。
solution1：(递归)

bool isMatch(const char *s, const char *p) {
    if (!*p)    
        return (!*s);

    if ('*' == *(p + 1)) {
        // x* matches empty string or at least one character: x* -> xx*
        // *s is to ensure s is non-empty
        return (isMatch(s, p + 2) || *s && (*s == *p || '.' == *p) && isMatch(s + 1, p));
    } 
    else {
        if (!*s)    
            return false;
        return (*s == *p || '.' == *p) ? isMatch(s + 1, p + 1) : false;
    }
}

solution2：(动态规划)

 bool isMatch(const char *s, const char *p) {
     int i, j;
     int m = strlen(s);
     int n = strlen(p);
     /**
      * 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 if 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];
     vector< vector<int> > b(m+1,vector<int>(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
     b[0][1] = false;
     for (j = 1; j < n; j++) {
         b[0][j + 1] = '*' == p[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] && ('.' == p[j] || s[i] == p[j]);
             } 
             else {
                 b[i + 1][j + 1] = b[i + 1][j - 1] && j > 0 || b[i + 1][j] ||
                     b[i][j + 1] && j > 0 && ('.' == p[j - 1] || s[i] == p[j - 1]);
             }
         }
     }
     return b[m][n];
 }

原文链接：https://oj.leetcode.com/discuss/18970/concise-recursive-and-dp-solutions-with-full-explanation-in