[LeetCode] 500. Keyboard Row 键盘行

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:

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.

给一组单词，返回哪些单词是由键盘上的一行字母组成的。

Java:

public class Solution {
    public String[] findWords(String[] words) {
        String[] strs = {"QWERTYUIOP","ASDFGHJKL","ZXCVBNM"};
        Map<Character, Integer> map = new HashMap<>();
        for(int i = 0; i<strs.length; i++){
            for(char c: strs[i].toCharArray()){
                map.put(c, i);//put <char, rowIndex> pair into the map
            }
        }
        List<String> res = new LinkedList<>();
        for(String w: words){
            if(w.equals("")) continue;
            int index = map.get(w.toUpperCase().charAt(0));
            for(char c: w.toUpperCase().toCharArray()){
                if(map.get(c)!=index){
                    index = -1; //don't need a boolean flag. 
                    break;
                }
            }
            if(index!=-1) res.add(w);//if index != -1, this is a valid string
        }
        return res.toArray(new String[0]);
    }
}

Java: 1-Line Solution via Regex and Stream

public String[] findWords(String[] words) {
    return Stream.of(words).filter(s -> s.toLowerCase().matches("[qwertyuiop]*|[asdfghjkl]*|[zxcvbnm]*")).toArray(String[]::new);
}

Python: wo

class Solution(object):
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        rows = {'q': 1, 'w': 1, 'e': 1, 'r': 1, 't': 1, 'y': 1, 'u': 1, 'i': 1, \
                'o': 1, 'p': 1, 'a': 2, 's': 2, 'd': 2, 'f': 2, 'g': 2, 'h': 2, \
                'j': 2, 'k': 2, 'l': 2, 'z': 3, 'x': 3, 'c': 3, 'v': 3, 'b': 3, \
                'n': 3, 'm': 3}
        res = []
        for word in words:
            row = None
            one_line = True
            for c in word:
                if not row:
                    row = rows[c.lower()] 
                elif row != rows[c.lower()]:
                    one_line = False
                    break
            if one_line:        
                res.append(word)   
                               
        return res 

Python:

def findWords(self, words):
    line1, line2, line3 = set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm')
    ret = []
    for word in words:
      w = set(word.lower())
      if w.issubset(line1) or w.issubset(line2) or w.issubset(line3):
        ret.append(word)
    return ret　　

C++:

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string> res;
        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'};
        for (string word : words) {
            int one = 0, two = 0, three = 0;
            for (char c : word) {
                if (c < 'a') c += 32;
                if (row1.count(c)) one = 1;
                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;
    }
};

　　

 

 

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