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.

American keyboard

Example 1:

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

Note:

You may use one character in the keyboard more than once.
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.

 1 class Solution {
 2 public:
 3     vector<string> findWords(vector<string>& words) {
 4         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},
 5                                    {'A',1},{'S',1},{'D',1},{'F',1},{'G',1},{'H',1},{'J',1},{'K',1},{'L',1},
 6                                    {'Z',2},{'X',2},{'C',2},{'V',2},{'B',2},{'N',2},{'M',2}};
 7         vector<string> res;
 8         for (string word: words){
 9             int i;
10             char firstletter=word[0]<='z'&&word[0]>='a'?word[0]-32:word[0];
11             for (i=1;i<word.length();i++){
12                 if(m[word[i]<='z'&&word[i]>='a'?word[i]-32:word[i]]!=m[firstletter]){
13                     break;
14                 }
15             }
16             if (i==word.length()){
17                 res.push_back(word);
18             }
19         }
20         return res;
21     }
22 };

Solution 2: use unordered_set

 1 class Solution {
 2 public:
 3     vector<string> findWords(vector<string>& words) {
 4         unordered_set<char> row1={'Q','W','E','R','T','Y','U','I','O','P'};
 5         unordered_set<char> row2={'A','S','D','F','G','H','J','K','L'};
 6         unordered_set<char> row3={'Z','X','C','V','B','N','M'};
 7         vector<string> res;
 8         for(string word:words){
 9             int one=0,two=0,three=0;
10             for(char c:word){
11                 c=(c>='a')?c-32:c;
12                 if (row1.count(c)) one = 1;//can't use one=row1.count(c),because we need to save the previous value of one
13                 if (row2.count(c)) two = 1;
14                 if (row3.count(c)) three = 1;
15                 if (one+two+three>1) break;
16             }
17             if (one+two+three==1) res.push_back(word);
18         }
19         return res;
20     }
21 };