500. Keyboard line

1. Title Description

Here's an array of strings words , Returns only those that can be used in American keyboard Words printed with letters on the same line . The keyboard is shown in the figure below .

American keyboard in ：

The first line consists of the characters “qwertyuiop” form .
The second line consists of characters “asdfghjkl” form .
The third line consists of the characters “zxcvbnm” form .

Example ：

Input ：words = [“Hello”,“Alaska”,“Dad”,“Peace”]
Output ：[“Alaska”,“Dad”]
Input ：words = [“adsdf”,“sfd”]
Output ：[“adsdf”,“sfd”]

2. Their thinking

For the input words , Convert it to lowercase first , Then judge whether each letter of the word is a subset of a certain line .

• stay python Through aggregate To quickly determine whether the letters that make up the word are a subset of a line

class Solution:
    def findWords(self, words: List[str]) -> List[str]:
    	line1 = set('qwertyuiop')
    	line2 = set('asdfghjkl')
    	line3 = set('zxcvbnm')
    	def f(x):
    		x = set(x.lower())
    		return x.issubset(line1) or x.issubset(line2) or x.issubset(line3)
    	return list(filter(f, words))

• stay C++ Judge whether each letter of the word is in a line in turn

class Solution {
    
public:
    vector<string> findWords(vector<string>& words) {
    
        vector<string> res;
        string line1 = "qwertyuiopQWERTYUIOP";
        string line2 = "asdfghjklASDFGHJKL"; 
        string line3 = "zxcvbnmZXCVBNM";
        for(int i=0;i<words.size();i++) {
    
            int n1 = 0, n2 = 0, n3 = 0;
            int length = words[i].length();
            for(int j=0;j<length;j++) {
    
                if(line1.find(words[i][j]) != string::npos) {
    
                    n1++;
                }else if(line2.find(words[i][j]) != string::npos) {
    
                    n2++;
                }else if(line3.find(words[i][j]) != string::npos) {
    
                    n3++;
                }
            }
            if(n1 == length || n2 == length || n3 == length) {
    
                res.push_back(words[i]);
            }
        }
        return res;
    }
};

3. Knowledge review

• python– aggregate
• python–filter function