问题描述：
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.
思路：
这道题思路很直接，就是作字符映射，一个字符映射成另一个字符，而且不能有两个不同的字符映射到同一个字符。
代码中用replace数组来存放字符映射表，而have_mapped数组记录某字符是否已经被映射，避免被两个不同的字符映射。
代码：
class Solution {
public:bool isIsomorphic(string s, string t) {char replace[150];bool have_mapped[150];for(int i = 0; i < 150; i++){replace[i] = 0;have_mapped[i] = false; //avoid the situation that two characters map to the same character}for(int i = 0; i < s.size();i++){int s_index = s[i] - 0;int t_index = t[i] - 0;if(replace[s_index] == 0) {if(have_mapped[t_index] == true) return false; //test if two characters map to the same characterreplace[s_index] = t[i];have_mapped[t_index] = true;}else{if(replace[s_index] != t[i]) return false;}}return true;}
};