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

標籤：style   blog   ar   io   color   sp   for   on   div   

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

 

我的思路：不提了，太挫了，寫了100多行代碼都沒搞定，直接看大神10行搞定的代碼吧：

其實主要的問題就在於p中的*究竟代表哪幾個字母，大神的代碼中用ss記錄*代表字母的後面一個位置，star記錄p中*的位置。

先假設*代表0個字元，如果後面發現不成立，再返回來

設*代表1個字元，............................

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;      }

 

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