題目來源
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.
然后我想了一個非常弱的解決方案…遍歷，兩個哈希表來記錄s到t，t到s的映射關系，假如關系不對，就返回false。雖然A了，但是還是有點亂…再想想。

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        int n = s.size();
        unordered_map<char, char> map1;
        unordered_map<char, char> map2;
        for (int i=0; i<n; i++) {
            if (!map1[s[i]] && !map2[t[i]]) {
                map1[s[i]] = t[i];
                map2[t[i]] = s[i];
            }
            else {
                if (map1[s[i]] != t[i] && map2[t[i]] != s[i])
                    return false;
            }
        }
        return true;
    }
};

然后實際上只需要記錄當前字母前一次出現的位置就可以了，代碼如下：

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        int n = s.size();
        vector<int> m1(256, 0);
        vector<int> m2(256, 0);
        for (int i=0; i<n; i++) {
            if (m1[s[i]] != m2[t[i]])
                return false;
            m1[s[i]] = i + 1;
            m2[t[i]] = i + 1;
        }
        return true;
    }
};