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.

判断一个字符串的所有字母是否在键盘的同一行

C++(3ms):

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<unordered_set<char>> row  {row1,row2,row3} ;
        vector<string> res ;
        for(string word : words){
            int r = -1 ;
            for(int i = 0 ; i < 3 ; i++){
                if (row[i].count(tolower(word[0])))
                    r = i ;
            }
            if (r == -1)
                continue ;
            bool flag = true ;
            for(int j = 1 ; j < word.size() ; j++){
                if (row[r].count(tolower(word[j])) == 0){
                    flag = false ;
                    break ;
                }
            }
            if (flag)
                res.push_back(word) ;
        }
        return res ;
    }
};

Java(4ms):

class Solution {
    public String[] findWords(String[] words) {
        String[] strs = {"qwertyuiop","asdfghjkl","zxcvbnm"} ;
        Map<Character,Integer> map = new HashMap<>() ;
        List<String> res = new LinkedList<>() ;
        for (int i = 0 ; i < strs.length ; i++){
            for (char c : strs[i].toCharArray()){
                map.put(c,i) ;
            }
        }
        for(String word : words){
            if (word.equals("")) continue ;
            int index = map.get(word.toLowerCase().charAt(0)) ;
            for(char c : word.toLowerCase().toCharArray()){
                if (map.get(c) != index){
                    index = -1 ;
                    break ;
                }
            }
            if (index != -1)
                res.add(word) ;
        }
        return res.toArray(new String[res.size()]) ;
    }
}