Leetcode 10 - Regular Expression Matching

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

分两种情况讨论：
1. 如果p[i+1] == ‘*’，循环递归直到p[i]和s[i]无法匹配为止。
2. 如果p[i+1] != ‘*’，则判断p[i]和s[i]是否匹配，并递归。

class Solution {public:    bool isMatch(string s, string p) {        if(p=="") return s=="";        int i=0,j=0;        if(p[i+1]=='*'){            while(p[i]=='.'&&s[j]!='\0' || p[i]==s[j]){                if(isMatch(s.substr(j),p.substr(i+2))) return true;                j++;            }            return isMatch(s.substr(j),p.substr(i+2));        }else if(p[i]=='.'&&s[j]!='\0' || p[i]==s[j]){                return isMatch(s.substr(j+1),p.substr(i+1));            }        return false;        }};