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题目概述：
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.我的代码：
bool isIsomorphic(char* s, char* t) { int ls,lt; //字符串长度 int i,j; int nums[256]={0}; int numt[256]={0}; ls = strlen(s); lt = strlen(t); if(ls!=lt) return false; for(i=0; i C++推荐代码：
参考： 题目很简单，也很容易想到方法，就是记录遍历s的每一个字母，并且记录s[i]到t[i]的映射，当发现与已有的映射不同时，说明无法同构，直接return false。但是这样只能保证从s到t的映射，不能保证从t到s的映射，所以交换s与t的位置再重来一遍上述的遍历就OK了。
class Solution {public: bool isIsomorphic(string s, string t) { if (s.length() != t.length()) return false; map mp; for (int i = 0; i < s.length(); ++i) { if (mp.find(s[i]) == mp.end()) mp[s[i]] = t[i]; else if (mp[s[i]] != t[i]) return false; } mp.clear(); for (int i = 0; i < s.length(); ++i) { if (mp.find(t[i]) == mp.end()) mp[t[i]] = s[i]; else if (mp[t[i]] != s[i]) return false; } return true; }}; (By:Eastmount 2015-9-21 凌晨1点半 ) 转载地址：http://cut.baihongyu.com/