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  • [LeetCode] Wildcard Matching

    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
    

    From Other:

    class Solution {
        public:
            bool isMatch(const char *s, const char *p) {
                if (!*p && !*s) return true; // both empty, so sad but true
                if (*p == *s) return isMatch(s + 1, p + 1); // match!
                else if (*p == '?' && *s) return isMatch(s + 1, p + 1); // weird match!
                else if (*p == '*') {
                    int ret = false;
                    while (*p == '*') ++p; // I only need just one starlet ;)
                    if (!*p) return true; // ends with star, the Universe can fit into it now!
                    while (*s) { // brute force match
                        const char *ts = s, *tp = p;
                        while ((*ts == *tp || *tp == '?') && *ts) {
                            if (*tp == '*') break;
                            ++ts; ++tp;
                        }
                        if (!*ts && !*tp) return true; // both empty
                        // *tp is * then ok, otherwise no exact match :(
                        if (*tp == '*') { 
                            // we don't need to concern ourself with more exact matches as the * would take care of it, 
                            // and for rest brute force matching will be done
                            ret |= isMatch(ts, tp);
                            return ret;
                        }
                        if (!*ts) return false; // search exhausted yo! p has more than s can handle :O
                        ++s;
                    }
                    return ret;
                } else return false; // WAT
            }
        };
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  • 原文地址:https://www.cnblogs.com/Xylophone/p/3911877.html
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