leetcode 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.

解法1:

class Solution(object):
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        return [w for w in words if self.is_in_keyboard_row(w)]
    
    def is_in_keyboard_row(self, word):
        s1 = set("qwertyuiop")
        s2 = set("asdfghjkl")
        s3 = set("zxcvbnm")
        s = s1
        word = word.lower()
        if word[0] in s1:
            s = s1
        elif word[0] in s2:
            s = s2
        else:
            s = s3
        i = 1
        while i < len(word):
            if word[i] not in s:
                return False
            i += 1
        return True

代码重构，像if比较多的可以用查找表做。set可以使用26数组来表示map替代。

解法2:

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

直接使用python的subset函数。

因为：

s.issubset(t)  测试是否 s 中的每一个元素都在 t 中

解法3:

使用正则表达式

def findWords(self, words):
    return filter(re.compile('(?i)([qwertyuiop]*|[asdfghjkl]*|[zxcvbnm]*)$').match, words)

解法4:

利用位运算省掉一开始找单词首字母在哪个set：

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<int> dict(26);
        vector<string> rows = {"QWERTYUIOP", "ASDFGHJKL", "ZXCVBNM"};
        for (int i = 0; i < rows.size(); i++) {
            for (auto c : rows[i]) dict[c-'A'] = 1 << i;
        }
        vector<string> res;
        for (auto w : words) {
            int r = 7;
            for (char c : w) {
                r &= dict[toupper(c)-'A'];
                if (r == 0) break;
            }
            if (r) res.push_back(w);
        }
        return res;
    }
};