### 422. Valid Word Square 题目: 难度 : Easy 思路: 就是对比一个矩阵内 xy == yx? try /except 真是好用 AC代码 ``` class Solution(object): def validWordSquare(self, words): """ :type words: List[str] :rtype: bool """ n = len(words) for i in xrange(n): m = len(words[i]) for j in xrange(m): try: if words[i][j] != words[j][i]: return False except: return False return True ```