500. Keyboard Row

其实这种问题的通用解法是动态规划，但是我的水平不够所以还写不出来。

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

解1：

class Solution(object):    def findWords(self, words):        """        :type words: List[str]        :rtype: List[str]        """        rows = ["qwertyuiop", "asdfghjkl","zxcvbnm"]        output = []        for word in words:            #word = word.lower()            for row in rows:                #row = row.lower()                num = 0                for char in word.lower():                    for row_char in row.lower():                        if char == row_char:                            num += 1                            break #这里加个这个，匹配到一个以后推出本次循环                if num == len(word):                    output.append(word)                    break        return output

这种应该是最笨的方法了，其中需要把判断的字符都转换为小写（string.lower() ），排名在34%。

解2：
还可以用set()来解，因为set()是所有出现过的字符，如果两个集合的交集等于单词的集合，那么这两个字符就匹配。

    def findWords(self, words):        """        :type words: List[str]        :rtype: List[str]        """        a=set('qwertyuiop')        b=set('asdfghjkl')        c=set('zxcvbnm')        ans=[]        for word in words:            t=set(word.lower())            if a&t==t:                ans.append(word)            if b&t==t:                ans.append(word)            if c&t==t:                ans.append(word)        return ans