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
正则表达式的匹配，注意*代表的不是任意的字符，而是0个或者任意多个前向的字符，代码如下：
1 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 if(p.size() == 0) return s.size() == 0; 5 if(p[1] != ‘*‘){ 6 if(s.size() != 0 && (p[0] == s[0] || p[0] == ‘.‘)) 7 return isMatch(s.substr(1), p.substr(1)); 8 return false; 9 }else{ 10 while(s.size() != 0 &&(s[0] == p[0] || p[0] == ‘.‘)){//模式串匹配0个或者更多的字符 11 if(isMatch(s, p.substr(2))) 12 return true; 13 s = s.substr(1); 14 } 15 return isMatch(s, p.substr(2)); 16 } 17 } 18 };
LeetCode OJ：Regular Expression Matching（正则表达式匹配）
原文：http://www.cnblogs.com/-wang-cheng/p/4987594.html