【python3】leetcode 500. Keyboard Row （easy）_python keyboard row-优快云博客

本文链接：https://blog.youkuaiyun.com/maotianyi941005/article/details/86486799

本文介绍了一种算法，用于从给定的单词列表中筛选出那些仅使用键盘上某一行字母组成的单词。通过将单词转换为小写并利用集合运算，可以高效地找出符合条件的单词。文章还提供了一个简洁的解决方案，利用Python的all和any函数进一步优化了筛选过程。

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500. Keyboard Row （easy）

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:

You may use one character in the keyboard more than once.
You may assume the input string will only contain letters of alphabet.

1 my solution
思路：如果一个单词的字母全在键盘的某一行，那个这个单词的set和某一行的交集就是这个set的length

class Solution:
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """

        if not words:return []
        line_1 = "qwertyueiop"
        line_2 = "asdfghjkl"
        line_3 = "zxcvbnm"
        row = []
        for word in words:
            wordset = set(word.lower())
            length = len(wordset)
            if len(wordset.intersection(line_1))== length or len(wordset.intersection(line_2)) == length or len(wordset.intersection(line_3)) == length:
                row.append(word)
        return row

2 记录一个discussion里的solution

使用all和any 感觉还蛮有趣的

all(iterable)："有‘假’为False，全‘真’为True，iterable为空是True"
any(iterable)："有‘真’为True，全‘假’为False，iterable为空是False"

class Solution:
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        return [word for word in words if all([all([char in line for char in word.lower()])==any([char in line for char in word.lower()]) for line in ['qwertyuiop', 'asdfghjkl', 'zxcvbnm']])]