题目在这
不需要使用两个哈希表就可以实现，具体思路是使用pattern中的值作为key，使用str中的值作为value，然后进行判断处理

swift实现

//: Playground - noun: a place where people can play
func WordPattern(_ pattern : String, _ str : String) -> Bool{
    let strArr = str.split(separator: " ")
    var dic = [Character : String]()
  
    if strArr.count != pattern.characters.count || Set(strArr).count != Set(pattern.characters).count{//"abba","mu mu mu mu"
        return false
    }
    
    for (i,v) in pattern.characters.enumerated(){
        if dic[v] == nil{
            dic[v] = String(strArr[i])
        }else{
            if dic[v] != String(strArr[i]) {
                return false
            }
        }
    }
    return true
}
WordPattern("aass", "mu mu dy dy")

python实现

#-*- coding:utf-8 -*-
#@Filename:leetcode290
#@Date:2018-09-10 20:44
#@auther:Mudy
#@E-mail:2808050775@qq.com
#@Blog:txmudy.cn

class Solution(object):
    def wordPattern(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        dic = {}
        strArr = str.split(" ")
        if len(strArr) != len(pattern) or  len(set(strArr)) != len(set(pattern)):# "abba","mu mu mu mu"
            return False

        for i, v in enumerate(pattern):
            if v not in dic:
                dic[v] = strArr[i]
            else:
                if  dic[v] != strArr[i] :
                    print(dic[v],strArr[i],end="\n")
                    return False
        return True



if __name__ == "__main__":
    s = Solution()
    print(s.wordPattern("aabb","mudy mudy mudy mudy"))