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

思路：只有当模式串和字符串同时等于\0，才可以认为两个串匹配。在匹配中，对于每个位的匹配可以分为三种情况1、（相应位匹配||模式串为.&&字符串不是\0）&&模式串下一位是*2、（相应位匹配||模式串为.&&字符串不是\0）&&模式串下一位不是*3、相应位不匹配&&（模式位不为.||字符串是\0）对应1，最复杂。分为*取0，*取1，*>=2三种情况。*取0对应跳过当前匹配位，继续寻找patter的下一个匹配位，str不变，pattern+2*取1对应当前匹配位算一次成功匹配，str+1，pattern+2*取>=2对应一次成功匹配，继续匹配字符串的下一位是否匹配，str+1，pattern不变三者取或。即只要有一种情况能匹配成功认为字符串就是匹配成功的。对应2，相当于一次成功匹配，str+1，pattern+1对应3，匹配失败，直接返回false

bool isSubMatch(string& s, int pos1, string& p, int pos2){    if (pos1 == s.size() && pos2 == p.size())return true;    if (pos1 != s.size() && pos2 == p.size())return false;    if (p[pos2 + 1] == '*'){        if (s[pos1] == p[pos2] || (p[pos2] == '.' && pos1 != s.size())){            return isSubMatch(s, pos1, p, pos2 + 2) || isSubMatch(s, pos1 + 1, p, pos2);        }        else            return isSubMatch(s, pos1, p, pos2 + 2);    }    if (s[pos1] == p[pos2] || (p[pos2] == '.' && pos1 != s.size()))        return isSubMatch(s, pos1 + 1, p, pos2 + 1);    return false;}bool isMatch(string s, string p) {    if (s == "" || p == "")return false;    return isSubMatch(s, 0, p, 0);}