Wildcard Matching

时间复杂度O(mn)，当碰到*通配符时，记录当前的s和p的位置，并从s继续向后匹配，若中间匹配失败，则尝试讲s+1，然后与p匹配。

class Solution {public:    bool isMatch(const char *s, const char *p) {    const char *ss = 0, *pp = 0;    while(*s)    {    if(*s == *p || *p == '?')    s++, p++;    else if(*p == '*')    {    while(*p == '*') p++;    if(*p == 0) return true;    ss = s;    pp = p;    }    else if(ss)    {    s = ++ss;    p = pp;    }    else    return false;    }    while(*p == '*') p++;    return *p == 0 && *s == 0;    }};