README_EN.md

500. Keyboard Row

中文文档

Description

Given an array of strings words, return the words that can be typed using letters of the alphabet on only one row of American keyboard like the image below.

In the American keyboard:

• the first row consists of the characters "qwertyuiop",
• the second row consists of the characters "asdfghjkl", and
• the third row consists of the characters "zxcvbnm".

 

Example 1:

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

Example 2:

Input: words = ["omk"]
Output: []

Example 3:

Input: words = ["adsdf","sfd"]
Output: ["adsdf","sfd"]

 

Constraints:

• 1 <= words.length <= 20
• 1 <= words[i].length <= 100
• words[i] consists of English letters (both lowercase and uppercase). 

Solutions

Python3

class Solution:
    def findWords(self, words: List[str]) -> List[str]:
        s1 = set('qwertyuiop')
        s2 = set('asdfghjkl')
        s3 = set('zxcvbnm')
        res = []
        for word in words:
            t = set(word.lower())
            if t <= s1 or t <= s2 or t <= s3:
                res.append(word)
        return res

Java

class Solution {
    public String[] findWords(String[] words) {
        String s1 = "qwertyuiopQWERTYUIOP";
        String s2 = "asdfghjklASDFGHJKL";
        String s3 = "zxcvbnmZXCVBNM";
        List<String> res = new ArrayList<>();
        for (String word : words) {
            int n1 = 0, n2 = 0, n3 = 0;
            int n = word.length();
            for (int i = 0; i < n; ++i) {
                if (s1.contains(String.valueOf(word.charAt(i)))) {
                    ++n1;
                } else if (s2.contains(String.valueOf(word.charAt(i)))) {
                    ++n2;
                } else {
                    ++n3;
                }
            }
            if (n1 == n || n2 == n || n3 == n) {
                res.add(word);
            }
        }
        return res.toArray(new String[0]);
    }
}

class Solution {
    public String[] findWords(String[] words) {
        String s = "12210111011122000010020202";
        List<String> res = new ArrayList<>();
        for (String word : words) {
            Set<Character> t = new HashSet<>();
            for (char c : word.toLowerCase().toCharArray()) {
                t.add(s.charAt(c - 'a'));
            }
            if (t.size() == 1) {
                res.add(word);
            }
        }
        return res.toArray(new String[0]);
    }
}

C++

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        string s = "12210111011122000010020202";
        vector<string> ans;
        for (auto& word : words)
        {
            unordered_set<char> t;
            for (char c : word) t.insert(s[tolower(c) - 'a']);
            if (t.size() == 1) ans.push_back(word);
        }
        return ans;
    }
};

Go

func findWords(words []string) []string {
	s := "12210111011122000010020202"
	var ans []string
	for _, word := range words {
		t := make(map[byte]bool)
		for _, c := range word {
			t[s[unicode.ToLower(c)-'a']] = true
		}
		if len(t) == 1 {
			ans = append(ans, word)
		}
	}
	return ans
}

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