Isomorphic String

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.

 

Analyse:

1. map twice.

    Runtime: 68ms.

 1 class Solution {
 2 public:
 3     bool isIsomorphic(string s, string t) {
 4         if(s.length() == 0) return true;
 5         
 6         map<char, char> m;
 7         map<char, char> n;
 8         for(int i = 0; i < s.length(); i++){
 9             if(m.find(s[i]) == m.end() && n.find(t[i]) == n.end()){
10                 m[s[i]] = t[i];
11                 n[t[i]] = s[i];
12             }
13             else
14                 if(m[s[i]] != t[i] || n[t[i]] != s[i]) return false;
15         }
16         return true;
17     }
18 };
View Code

 

　 Runtime: 104ms.

 1 class Solution {
 2 public:
 3     bool isIsomorphic(string s, string t) {
 4         if(s.length() == 0) return true;
 5         
 6         map<char, char> m;
 7         m[s[0]] = t[0];
 8         for(int i = 1; i < s.length(); i++){
 9             if(m.find(s[i]) != m.end() && m.find(s[i])->second != t[i]) return false;
10             else
11                 m.insert(pair<char, char> (s[i], t[i]));
12         }
13         map<char, char> n;
14         n[t[0]] = s[0];
15         for(int i = 1; i < s.length(); i++){
16             if(n.find(t[i]) != n.end() && n.find(t[i])->second != s[i]) return false;
17             else
18                 n.insert(pair<char, char> (t[i], s[i]));
19         }
20         return true;
21     }
22 };
View Code

 

　　Runtime: 140ms

 1 class Solution {
 2 public:
 3     bool isIsomorphic(string s, string t) {
 4         if(s.length() == 0) return true;
 5         
 6         map<char, char> m;
 7         m[s[0]] = t[0];
 8         map<char, char> n;
 9         n[t[0]] = s[0];
10         for(int i = 1; i < s.length(); i++){
11             if(m.find(s[i]) != m.end() && m.find(s[i])->second != t[i] ||
12                n.find(t[i]) != n.end() && n.find(t[i])->second != s[i]) return false;
13             else{
14                 m.insert(pair<char, char> (s[i], t[i]));
15                 n.insert(pair<char, char> (t[i], s[i]));
16             }
17         }
18         return true;
19     }
20 };
View Code

 

posted @ 2015-08-19 00:47  amazingzoe  阅读(188)  评论(0)  编辑  收藏  举报