题目如下：

解题思路：本题的关键是pattern和word之间必须是一对一的关系。因此需要建立pattern->word和word->pattern两种映射，这两种映射可用两个字典分别保存。

代码如下：

class Solution(object):    def wordPattern(self, pattern, str):        """        :type pattern: str        :type str: str        :rtype: bool        """        words = str.split(' ')        if len(pattern) != len(words):            return False        dic_p_to_w = {}        dic_w_to_p = {}        for p,w in zip(pattern,words):            if p not in dic_p_to_w and w not in dic_w_to_p:                dic_p_to_w[p] = w                dic_w_to_p[w] = p            elif p in dic_p_to_w and w in dic_w_to_p:                if (dic_p_to_w[p] == w and dic_w_to_p[w] == p) == False:                    return False            else:                return False        return True