LeetCode之“散列表”：Isomorphic Strings

　　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 }