标签：false nbsp code string i++ aced etc iso 多对一
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.
设定两个字符串长度相等，所谓的“同构”，就是字符串 s 中的字符可以一对一的映射到字符串 t 中的字符。不能一对多，也不能多对一。
利用两个map实现。
代码如下：
1 class Solution { 2 public: 3 bool isIsomorphic(string s, string t) { 4 map<char, char> mapA; 5 map<char, char> mapB; 6 int n = s.length(); 7 for(int i = 0; i < n; i++) 8 { 9 char ss = s[i]; 10 char tt = t[i]; 11 map<char, char>::iterator it = mapA.find(ss); 12 if(it != mapA.end()) 13 { 14 if(mapA[ss] != tt) 15 { 16 return false; 17 } 18 } 19 else 20 { 21 map<char, char>::iterator ii = mapB.find(tt); 22 if(ii != mapB.end() && mapB[tt] != ss) 23 { 24 return false; 25 } 26 else 27 { 28 mapA[ss] = tt; 29 mapB[tt] = ss; 30 } 31 } 32 } 33 return true; 34 } 35 };
标签：false nbsp code string i++ aced etc iso 多对一
原文地址：http://www.cnblogs.com/shellfishsplace/p/6050946.html
