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

 

首先这里有个可能大家不知道的地方：

if p[0] = '*', the string must be an invalid string.

that is what Regular Expression defined.

这是别人在评论的时候说的。也就是说，pattern的第一个不可以是*；

 

class Solution {public:    bool isMatch(string s, string p) { if (p.empty()) return s.empty(); 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.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p)); else return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1)); } }; class Solution { public: bool isMatch(string s, string p) { /** * f[i][j]: if s[0..i-1] matches p[0..j-1] * if p[j - 1] != '*' * f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1] * if p[j - 1] == '*', denote p[j - 2] with x * f[i][j] is true iff any of the following is true * 1) "x*" repeats 0 time and matches empty: f[i][j - 2] * 2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j] * '.' matches any single character */ int m = s.size(), n = p.size(); vector
     
      
       > f(m + 1, vector
       
        (n + 1, false)); f[0][0] = true; for (int i = 1; i <= m; i++) f[i][0] = false; // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty for (int j = 1; j <= n; j++) f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2]; for (int i = 1; i <= m; i++) for (int j = 1; j <= n; j++) if (p[j - 1] != '*') f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]); else // p[0] cannot be '*' so no need to check "j > 1" here f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j]; return f[m][n]; } };