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]]=i
                continue
            #记录其他下标
            dit1[s[i]]+=i
        for j in range(len(t)):
            if t[j] not in dit2:
                dit2[t[j]]=j
                continue
             #记录其他下标
            dit2[t[j]]+=j
        ret1=list(dit1.values())
        ret2=list(dit2.values())
        if ret1==ret2:
            return True
        return 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 False
    
        for i in range(len(s)):
            inds = list(dicts.keys()).index(s[i])
            if list(dictt.keys())[inds] != t[i]:
                return False
        return True