给定两个字符串，检查它们是否是结构相似的。

对于结构相似的定义如下

Two strings s and t are isomorphic if the characters in s can be replaced to get t.
All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.

两个字符串的每个字符之间存在某种唯一确定的映射关系，且必须是一一对应，即不能有两个不同的字符对应同一个字符。同时该映射可以是自己到自己。

我的直接想法就是直接建立两组映射关系，即从s->t 和 t->s。然后去check是否一一对应。代码如下

from collections import defaultdict
class Solution1(object):
    def isIsomorphic(self, s, t):
        if len(s)!=len(t):
            return False
        else:
            mapp=defaultdict(list)
            rev_mapp=defaultdict(list)
            
            for i,j in zip(s,t):
                if j not in mapp[i]:
                    mapp[i].append(j)
                if i not in rev_mapp[j]:
                    rev_mapp[j].append(i)
            
            for k,v in mapp.items():
                if len(v)>1:
                    return False
            for k,v in rev_mapp.items():
                if len(v)>1:
                    return False
            return True

对于字符串s和t,都去建立一个转换字符，类似于数字编码的过程。然后去比较两个字符串的编码是否相同。转换字符的过程通过transformString函数来实现。代码如下

class Solution2(object):
    def transformString(self,s):
        idx_mapping={}
        new_str=[]

        for i,item in enumerate(s):
            if item not in idx_mapping:
                idx_mapping[item]=i
            new_str.append(str(idx_mapping[item]))
        return ' '.join(new_str)
    
    def isIsomorphic(self,s,t):
        return self.transformString(s)==self.transformString(t)