题目：
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"]题意：
给出n个字符串，从而判断每个字符串中的字符石头来自美式键盘上的同一行，若来自同一行，返回该string。
代码：
class Solution(object): def findWords(self, words): """ :type words: List[str] :rtype: List[str] """ dict = {'Q':1, 'W':1, 'E':1, 'R':1, 'T':1, 'Y':1, 'U':1, 'I':1, 'O':1, 'P':1, 'A':2, 'S':2, 'D':2, 'F':2, 'G':2, 'H':2, 'J':2, 'K':2, 'L':2, 'Z':3, 'X':3, 'C':3, 'V':3, 'B':3, 'N':3, 'M':3, } //定义一个字典 n = len(words) //统计字符串长度 i = 0 while i < n : flag = 1 ni = len(words[i]) if ni > 0 : if ord(words[i][0]) <=122 and ord(words[i][0]) >= 97 : //小写字符的处理 x = dict[chr(ord(words[i][0])-32)] else : x = dict[words[i][0]] j = 1 //比较字符串里的每个字符 while j < ni : if ord(words[i][j]) <=122 and ord(words[i][j]) >= 97 : //小写字符处理 y = dict[chr(ord(words[i][j])-32)] else : y = dict[words[i][j]] if x != y : //如果有字符不在同一行，则删除该字符串 del words[i] flag = 0 break j += 1 if flag == 0 : n -= 1 i -= 1 i += 1 return words
网上其他的处理方法，充分利用python的子集功能：
class Solution2(object): def findWords(self, words): row1, row2, row3 = set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm'); ret = []; for word in words: w = set(word.lower()); if w.issubset(row1) or w.issubset(row2) or w.issubset(row3): ret.append(word); return ret;
厉害！！！
