思路：

先记录同一个值出现的次数，再将字典中的值取出，比较2个列表即可 

代码：

class Solution:def isIsomorphic(self, s: str, t: str) -> bool:dit1=dict()dit2=dict()for i in range(len(s)):if s[i] not in dit1:dit1[s[i]]=icontinue#记录其他下标dit1[s[i]]+=ifor j in range(len(t)):if t[j] not in dit2:dit2[t[j]]=jcontinue#记录其他下标dit2[t[j]]+=jret1=list(dit1.values())ret2=list(dit2.values())if ret1==ret2:return Truereturn False

还有高手？思路差不多，更精简 

class Solution:def isIsomorphic(self, s: str, t: str) -> bool:#记录值的出现次数dicts = Counter(s)dictt = Counter(t)        if list(dicts.values()) != list(dictt.values()):return Falsefor i in range(len(s)):inds = list(dicts.keys()).index(s[i])if list(dictt.keys())[inds] != t[i]:return Falsereturn True