890. Find and Replace Pattern

簡單來說, 就是用一個map 紀錄, 紀錄出現過的字母, 如果再重複被map access, 就檢查是不是原本檢查那個, 另外也要再用一個set紀錄被記錄過的人

class Solution {
public:
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> res;
        for(auto word: words){
            unordered_map<char, char> m;
            unordered_set<char> st;
            int i = 0;
            bool flag = true;
            for(; i < word.size();i++){
                if(!m.count(pattern[i])){
                    if(st.count(word[i])){
                        flag = false;
                        break;
                    }
                    m[pattern[i]] = word[i];
                    st.insert(word[i]);
                }else{
                    if(m[pattern[i]] != word[i]){
                        flag =false;
                        break;
                    };
                }
            }
            if(flag) res.push_back(word);
        }
        return res;
    }
};

李哥的做法果然還是精闢, 先是用一個map, 紀錄字母出現的順序, 例如

caa->122, 像這樣, 再把數字轉乘字母

122->abb

倘若兩邊轉出來的字串一樣, patternn就一樣了

class Solution {
public:
    string check(string s){
        unordered_map<char, int> m;
        for(auto c: s){
            if(!m.count(c)){
                m[c] =  m.size();
            }
        }
        string res;
        for(auto c: s){
            res += m[c] + 'a';
        }
        return res;
    }
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string > res;
        for(auto word: words){
            if(check(word) == check(pattern)) res.push_back(word);
        }
        return res;
    }
};