LeetCode 鍵盤行

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string> res;  //定義一個string類型的vector
        unordered_set<char> a{'q','w','e','r','t','y','u','i','o','p','Q','W','E','R','T','Y','U','I','O','P'};
        unordered_set<char> b{'A','S','D','F','G','H','J','K','L','a','s','d','f','g','h','j','k','l'};
        unordered_set<char>
            c{'z','x','c','v','b','n','m','Z','X','C','V','B','N','M'};
        //a,b,c爲無序集合，分別包含鍵盤每行的字母大小寫；
        
        for(int i=0;i!=words.size();i++){  //對words中的每一個字符串進行循環
            int x=0;
            int y=0;
            int z=0;//每次處理完一個字符串均要初始化xyz的值，防止後面的判斷出錯；
            for(int j=0;j!=words[i].size();j++) //對每個字符串中的字符進行循環
            {
              if(a.count(words[i][j])) x=1;
              else if(b.count(words[i][j])) y=1;
              else if(c.count(words[i][j])) z=1; //unordered.count(i)是返回i出現的次數，   若                                                                                    出現某一行鍵盤的字母，則設置x/y/z=1；
                if(x+y+z>1) break;            //若是出現超過一行的字母，則跳出，進行下一個字                                                                       符串的判斷；
            }
            if(x+y+z==1){res.push_back(words[i]);} //若該字符串僅含一行鍵盤的字母，則將其添                                                                              加至res中；
        }
        return res;
        
    }
};