【leetcode】290. Word Pattern

使用两个map，来分别存放一一对应。细节方面就是对于字符的切割，以及如何查询map的键值是否存在。

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        if(str.size() == 0 && pattern.size() == 0) return true;
        vector<string> vec;
        preprocess(str, vec);
        if(vec.size() != pattern.size()) return false;
        
        unordered_map<char, string> csmap;
        unordered_map<string, char> scmap;
        for(int i = 0; i < pattern.size(); ++ i) {
            string word = csmap[pattern[i]];
            if(word.size() == 0) {
                csmap[pattern[i]] = vec[i];
                if(scmap[vec[i]]) return false;
                scmap[vec[i]] = pattern[i];
            }
            else {
                if(word != vec[i]) return false;
            }
        }
        return true;
    }
    
private:
    void preprocess(string str, vector<string>& vec) {
        if(str.size() == 0) return;
        int i = 0;
        for(; i < str.size(); ++ i) {
            if(str[i] == ' ') break;
        }
        if(i == str.size())  vec.push_back(str);
        else {
            vec.push_back(str.substr(0, i));
            preprocess(str.substr(i+1), vec);
        }
    }
};

class Solution {
public:
    bool wordPattern(string pattern, string str) {
		if(pattern.empty() && str.empty()) return true;
		if(pattern.empty() || str.empty()) return false;
		vector<string> word;
		process(str, word);
		if(word.size() != pattern.size()) return false;
        
		unordered_map<char, string> cTos;
		unordered_map<string, char> sToc;
		for(int i = 0; i < pattern.size(); ++ i) {
			if(cTos.find(pattern[i]) == cTos.end() && sToc.find(word[i]) == sToc.end()) {
				cTos[pattern[i]] = word[i];
				sToc[word[i]] = pattern[i];
			}
			else if(cTos.find(pattern[i]) == cTos.end() || sToc.find(word[i]) == sToc.end()) 
				return false;
			else {
				if(cTos[pattern[i]] != word[i]) return false;
				if(sToc[word[i]] != pattern[i]) return false;
			}
		}
		return true;
	}

private:
	void process(string& s, vector<string>& word) {
		while(!s.empty()) {
			int pos = s.find(' ');
			if(pos == -1) {
				word.push_back(s);
				s.clear();
			}
			else {
				word.push_back(s.substr(0, pos));
				s = s.substr(pos + 1);
			}
		}
	}
};