【leetcode】Isomorphic Strings（easy)

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.

秒小题好爽的说。

思路：保存两个字符串字母对应的字母，出现不一致就false

我的代码：

bool isIsomorphic(string s, string t) {
        unordered_map<char, char> map1, map2;
        for(int i = ; i < s.size(); ++i)
        {
            if(map1.find(s[i]) == map1.end() && map2.find(t[i]) == map2.end())
            {
                map1[s[i]] = t[i]; map2[t[i]] = s[i];
            }
            else if(map1[s[i]] != t[i] || map2[t[i]] != s[i])
                return false;
            else;
        }
        return true;
    }

网上C版本代码 免去了查找 更快

bool isIsomorphic(char* s, char* t) {
        char mapST[] = {  };
        char mapTS[] = {  };
        size_t len = strlen(s);
        for (int i = ; i < len; ++i)
        {
            if (mapST[s[i]] ==  && mapTS[t[i]] == )
            {
                mapST[s[i]] = t[i];
                mapTS[t[i]] = s[i];
            }
            else
            {
                if (mapST[s[i]] != t[i] || mapTS[t[i]] != s[i])
                    return false;
            }
        }
        return true;
}