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  • regular-expression-matching

    https://leetcode.com/problems/regular-expression-matching/

    我觉得这个递归解法很好,简洁明了:

    public:
        bool isMatch(string s, string p) {
            if (p.empty())    return s.empty();
            
            if ('*' == p[1])
                // x* matches empty string or at least one character: x* -> xx*
                // *s is to ensure s is non-empty
                return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));
            else
                return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1));
        }
    };

    也可以方便的改成DP.

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  • 原文地址:https://www.cnblogs.com/charlesblc/p/6364704.html
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