500. Keyboard Row

Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below. 

Example 1:

Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Alaska", "Dad"]

Note:

1. You may use one character in the keyboard more than once.
2. You may assume the input string will only contain letters of alphabet.

Solution 1: use unordered_map, if each char's value in a word equal the value of the first char, the word can be typed using the letters in the same row.

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        unordered_map<char,int> m={{'Q',0},{'W',0},{'E',0},{'R',0},{'T',0},{'Y',0},{'U',0},{'I',0},{'O',0},{'P',0},
                                   {'A',1},{'S',1},{'D',1},{'F',1},{'G',1},{'H',1},{'J',1},{'K',1},{'L',1},
                                   {'Z',2},{'X',2},{'C',2},{'V',2},{'B',2},{'N',2},{'M',2}};
        vector<string> res;
        for (string word: words){
            int i;
            char firstletter=word[0]<='z'&&word[0]>='a'?word[0]-32:word[0];
            for (i=1;i<word.length();i++){
                if(m[word[i]<='z'&&word[i]>='a'?word[i]-32:word[i]]!=m[firstletter]){
                    break;
                }
            }
            if (i==word.length()){
                res.push_back(word);
            }
        }
        return res;
    }
};

Solution 2: use unordered_set

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        unordered_set<char> row1={'Q','W','E','R','T','Y','U','I','O','P'};
        unordered_set<char> row2={'A','S','D','F','G','H','J','K','L'};
        unordered_set<char> row3={'Z','X','C','V','B','N','M'};
        vector<string> res;
        for(string word:words){
            int one=0,two=0,three=0;
            for(char c:word){
                c=(c>='a')?c-32:c;
                if (row1.count(c)) one = 1;//can't use one=row1.count(c),because we need to save the previous value of one
                if (row2.count(c)) two = 1;
                if (row3.count(c)) three = 1;
                if (one+two+three>1) break;
            }
            if (one+two+three==1) res.push_back(word);
        }
        return res;
    }
};