【leetcode】Wildcard Matching（hard) ★ 大神太牛了

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

 

我的思路：不提了，太挫了，写了100多行代码都没搞定，直接看大神10行搞定的代码吧：

其实主要的问题就在于p中的*究竟代表哪几个字母，大神的代码中用ss记录*代表字母的后面一个位置，star记录p中*的位置。

先假设*代表0个字符，如果后面发现不成立，再返回来

设*代表1个字符，............................

bool isMatch(const char *s, const char *p) {
        const char* star=NULL;
        const char* ss=s;
        while (*s){
            //advancing both pointers when (both characters match) or ('?' found in pattern)
            //note that *p will not advance beyond its length 
            if ((*p=='?')||(*p==*s)){s++;p++;continue;} 

            // * found in pattern, track index of *, only advancing pattern pointer 
            if (*p=='*'){star=p++; ss=s;continue;} 

            //current characters didn't match, last pattern pointer was *, current pattern pointer is not *
            //only advancing pattern pointer
            if (star){ p = star+1; s=++ss;continue;} 

           //current pattern pointer is not star, last patter pointer was not *
           //characters do not match
            return false;
        }

       //check for remaining characters in pattern
        while (*p=='*'){p++;}

        return !*p;  
    }