【LeetCode】500. Keyboard Row【E】【75】

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.

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想法很简单 就是用集和运算 然后看每个word是不是一行的子集

最开始写的时候 太傻了 被注释掉了

class Solution(object):
    def findWords(self, words):
        #row1 = set(['Q','W','E','R','T','Y','U','I','O','P','q','w','e','r','t','y','u','i','o','p'])
        #row2 = set(['A','S','D','F','G','H','J','K','L','a','s','d','f','g','h','j','k','l'])
        #row3 = set(['Z','X','C','V','B','N','M','z','x','c','v','b','n','m'])
        row1 = set('qwertyuiop')
        row2 = set('asdfghjkl')
        row3 = set('zxcvbnm')

        res = []
        for i in words:
            si = set(i.lower())
            #if len(si - row1) == 0 or len(si - row2) == 0 or len(si - row3) == 0:
            if si.issubset(row1) or si.issubset(row2) or si.issubset(row3):
                res += i,
        return res
        """
        :type words: List[str]
        :rtype: List[str]
        """