[LeetCode]010-Regular Expression Matching

题目：
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

Solution:
//’.’很好匹配，’*’比较难匹配，采用递归算法
//(1)如果当前匹配的字符的后一位不是’*’,则直接对应匹配即可。（要考虑’.’)
//(2)如果当前匹配的字符的后一位是’*’,则分两步，一步是s不停后移与当前的p比较，直到出现不一样字符。一步是上一步结束后，将新的s与p+2(后移两位)开始重新比较

 bool isMatch(string s, string p)    {        if(p.size() == 0)            return s.size() == 0?true:false;        if(p.size() == 1)        {            if(s.size() == 1 && (p[0]=='.' || p[0] == s[0]))                return true;            else                return false;        }        if(p[1] != '*')        {            if(s.size()>0 && (p[0] == '.' || p[0] == s[0]))            {                return isMatch(s.substr(1),p.substr(1));            }            else                return false;        }        else        {            while(s.size() >0 && (s[0] == p[0] || p[0] == '.'))            {                if(isMatch(s,p.substr(2)))//aaa和a*a                    return true;                s = s.substr(1);            }            return isMatch(s,p.substr(2));        }    }