205 ISOMORPHIC STRINGS

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.

此题目有时间限制，关键是如何优化时间。

我开始的做法是两个for循环，那么时间复杂度就是n的平方，但是它有一个测试用例，两个字符串特别长，于是就出现了“Time Limit Exceeded”。代码如下：

class Solution {

public:

 bool isIsomorphic(string s, string t) {
    int len = s.length();
    // 时间复杂度n平方，不满足题目要求。
    for (size_t i = 0; i < len; i++) {
       for (size_t j = i + 1; j < s.length(); j++) {
          if ((s[i] == s[j] && t[i] != t[j]) || (s[i] != s[j] && t[i] == t[j])) {
              return false;
          }
       }
    }
    return true;
    }
};

上面的方法不行，那就必须要减少时间复杂度，最后我想了一个方法：使用一个<char, char>的map映射，for循环两个入参的每一个char，如果发现对应关系改变了，那么就说明两个字符串不是isomorphic的了。时间复杂度为O(n)，代码如下：

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        int len = s.length();
        map<char, char> m;
        map<char, char> m2;
        for (size_t i = 0; i < len; i++) {
            if (m.find(s[i]) == m.end()) {
                m[s[i]] = t[i];
            }else if (m[s[i]] != t[i]) {
                return false;
            }
            if (m2.find(t[i]) == m2.end()) {
                m2[t[i]] = s[i];
            }else if (m2[t[i]] != s[i]) {
                return false;
            }
        }
        return true;
    }
};