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  • 【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;  
        }
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  • 原文地址:https://www.cnblogs.com/dplearning/p/4172888.html
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