Leetcode isomorphic Strings

Given strings s and t, determine if they are isomorphic.

The strings is isomorphic if the characters in s can is replaced to get t.

All occurrences of a character must is replaced with another character while preserving the order of characters. No, characters may map to the same character and a character may map to itself.

For example,
Given "egg" , "add" return True.

Given "foo" , "bar" return FALSE.

Given "paper" , "title" return True.

Didn't do it.

Without a very clear understanding of the meaning of the topic, the topic is that two strings are isomorphic. I think it's the same string in the same position in T as long as the strings in the string s are the same.

In fact the constraint is more, it is a mapping, like egg,e and a mapping, G and D mapping, then the string s in the case of E, the string t must be a.

So, building a map, using the method you've seen last time, is OK. if (Map.find (S[i]) ==map.end ()) ****else****

However, there are two checks to check if the s->t mapping T is satisfied, and the t->s mapping S is satisfied.

1 classSolution {2  Public:3     BOOLIsisomorphic (stringSstringt) {4map<Char,Char>model;5         intSlength=s.length (), tlength=t.length ();6         if(slength!=tlength)return false;7          for(intI=0; i<slength;i++){8             if(Model.find (S[i]) ==model.end ()) model[s[i]]=T[i];9             Else if(Model[s[i]]!=t[i])return false;Ten         } One model.clear (); A          for(intI=0; i<slength;i++){ -             if(Model.find (T[i]) ==model.end ()) model[t[i]]=S[i]; -             Else if(Model[t[i]]!=s[i])return false; the         } -         return true; -  -          +     } -};

Leetcode isomorphic Strings