Isomorphic String

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

Two strings 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.

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

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:
You may assume both s and t have the same length.

Analyse:

1. map twice.

    Runtime: 68ms.

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        if(s.length() == 0) return true;
        
        map<char, char> m;
        map<char, char> n;
        for(int i = 0; i < s.length(); i++){
            if(m.find(s[i]) == m.end() && n.find(t[i]) == n.end()){
                m[s[i]] = t[i];
                n[t[i]] = s[i];
            }
            else
                if(m[s[i]] != t[i] || n[t[i]] != s[i]) return false;
        }
        return true;
    }
};

　 Runtime: 104ms.

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        if(s.length() == 0) return true;
        
        map<char, char> m;
        m[s[0]] = t[0];
        for(int i = 1; i < s.length(); i++){
            if(m.find(s[i]) != m.end() && m.find(s[i])->second != t[i]) return false;
            else
                m.insert(pair<char, char> (s[i], t[i]));
        }
        map<char, char> n;
        n[t[0]] = s[0];
        for(int i = 1; i < s.length(); i++){
            if(n.find(t[i]) != n.end() && n.find(t[i])->second != s[i]) return false;
            else
                n.insert(pair<char, char> (t[i], s[i]));
        }
        return true;
    }
};

　　Runtime: 140ms

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        if(s.length() == 0) return true;
        
        map<char, char> m;
        m[s[0]] = t[0];
        map<char, char> n;
        n[t[0]] = s[0];
        for(int i = 1; i < s.length(); i++){
            if(m.find(s[i]) != m.end() && m.find(s[i])->second != t[i] ||
               n.find(t[i]) != n.end() && n.find(t[i])->second != s[i]) return false;
            else{
                m.insert(pair<char, char> (s[i], t[i]));
                n.insert(pair<char, char> (t[i], s[i]));
            }
        }
        return true;
    }
};