题目要求：
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.
这题难度不大。
第一个程序（28ms）：
1 bool isIsomorphic(string s, string t) { 2 unordered_map<char, char> hashMap; 3 int sz = s.size(); 4 for(int i = 0; i < sz; i++) 5 { 6 if(hashMap.find(s[i]) != hashMap.end()) 7 { 8 if(hashMap[s[i]] != t[i]) 9 return false; 10 } 11 else 12 { 13 unordered_map<char, char>::iterator itr = hashMap.begin(); 14 for(; itr != hashMap.end(); itr++) 15 if(itr->second == t[i]) 16 return false; 17 18 hashMap[s[i]] = t[i]; 19 } 20 21 } 22 23 return true; 24 }
第二个程序（24ms）：
1 bool isIsomorphic(string s, string t) { 2 unordered_map<char, char> hashMap; 3 unordered_map<char, char> rHashMap; 4 int sz = s.size(); 5 for(int i = 0; i < sz; i++) 6 { 7 if(hashMap.find(s[i]) != hashMap.end()) 8 { 9 if(hashMap[s[i]] != t[i]) 10 return false; 11 } 12 else 13 { 14 if(rHashMap.find(t[i]) != rHashMap.end()) 15 return false; 16 17 hashMap[s[i]] = t[i]; 18 rHashMap[t[i]] = s[i]; 19 } 20 } 21 22 return true; 23 }
看了网上别人的做法（少掉了哈希表查找的时间损耗），只需8ms：
1 bool isIsomorphic(string s, string t) { 2 char map_s[128] = { 0 }; 3 char map_t[128] = { 0 }; 4 int len = s.size(); 5 for (int i = 0; i < len; ++i) 6 { 7 if (map_s[s[i]]!=map_t[t[i]]) return false; 8 map_s[s[i]] = i+1; 9 map_t[t[i]] = i+1; 10 } 11 12 return true; 13 }