[LeetCode291]Word Pattern II

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str.

Examples:
pattern = "abab", str = "redblueredblue" should return true.
pattern = "aaaa", str = "asdasdasdasd" should return true.
pattern = "aabb", str = "xyzabcxzyabc" should return false.
Notes:
You may assume both pattern and str contains only lowercase letters.

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這次的word pattern不是之前那個已經給我們match好的pattern,而是需要我們自己去match並且check是否存在valid pattern。

所以要backtracking驗證。 code 如下:

class Solution {
public:
    bool wordPatternMatch(string pattern, string str) {
        unordered_set<string> strMp;
        unordered_map<char, string> mp;
        return match(pattern, str, 0, 0, strMp, mp);
    }
private:
    bool match(string& pattern, string& str, int strPos, int patPos, unordered_set<string>& strMp, unordered_map<char, string>& p2s ){
        int m = str.size(), n = pattern.size();
        if(strPos == m && patPos == n) return true;
        if(n - patPos > m - strPos|| patPos == n) return false;// this condition reduce time!!
        char cur = pattern[patPos];
        // if cur char has matching string we try to verify it
        if(p2s.find(cur) != p2s.end()){
            string curStr = p2s[cur];
            int len = curStr.size();
            if(curStr != str.substr(strPos, len)) return false;
            return match(pattern, str, strPos + len, patPos + 1, strMp, p2s);
        }
        for(int i = strPos; i < m; ++i){
            string s = str.substr(strPos, i - strPos + 1);
            if(strMp.find(s) != strMp.end()) continue;
            strMp.insert(s);
            p2s[cur] = s;
            if(match(pattern, str, i+1, patPos+1, strMp, p2s)) return true;
            strMp.erase(s);
             p2s.erase(cur);
        }
        return false;
    }   
};
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