【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+; 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;

    }