[Leetcode 44, Hard] Wildcard match

最新推荐文章于 2024-06-22 09:48:08 发布

simplify2012

最新推荐文章于 2024-06-22 09:48:08 发布

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分类专栏： Leetcode_hard 文章标签： leetcode C++

本文链接：https://blog.youkuaiyun.com/u012212811/article/details/47079437

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本文介绍了如何实现一种通配符匹配算法，支持问号匹配任意单字符和星号匹配任意字符序列。通过C++、Java和Python三种语言的实例演示了算法的具体实现。

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Problem:

mplement 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

Analysis:

The solution comes from http://yucoding.blogspot.com/2013/02/leetcode-question-123-wildcard-matching.html

Solutions:

C++:

    bool isMatch(string S, string P) 
    {
        const char *s = S.c_str();
        const char *p = P.c_str();
        const char* star = NULL;
        const char* ss = s; 
        while (*s){
            if ((*p == '?')||(*p == *s)) {
                s++;
                p++;
                continue;
            }

            if (*p == '*') {
                star = p++; 
                ss = s;
                continue;
            }
            
            if (star) { 
                p = star+1; 
                s = ++ss;
                continue;
            }
            
            return false;
        }

        while (*p == '*')
            p++;

        return !*p;
    }

Java:

Python: