题目简述：

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.

解题思路:

首先，相同的地方必须相同这个条件可以得出一个O(n^2)的算法

class Solution:

    # @param {string} s

    # @param {string} t

    # @return {boolean}

    def isIsomorphic(self, s, t):

        ls = len(s)

        lt = len(t)

        if ls != lt:

            return False

        for i in range(ls):

            for j in range(ls):

                if s[i] == s[j]:

                    if t[i] != t[j]:

                        return False

        return True

但是很不幸超时了，于是我们用hash的方法得到另一个更快的算法：

class Solution:

    # @param {string} s

    # @param {string} t

    # @return {boolean}

    def isIsomorphic(self, s, t):

        ls = len(s)

        lt = len(t)

        if ls != lt:

            return False

        dic = {}

        dic2 = {}

        for i in range(ls):

            if s[i] not in dic.keys():

                dic[s[i]] = t[i]

            else:

                if dic[s[i]] != t[i]:

                    return False

            if t[i] not in dic2.keys():

                dic2[t[i]] = s[i]

            else:

                if dic2[t[i]] != s[i]:

                    return False

        return True