205. Isomorphic Strings

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.

解法一，使用map，时间复杂度O(nlgn)。

class Solution { public: bool isIsomorphic(string s, string t) { /* if(s.length() != t.length()) return false; map<char, char> mp1; map<char, char> mp2; for(int i=0; i<s.length(); ++i){ auto it1 = mp1.insert(make_pair(s[i], t[i])); if(!it1.second){ if(it1.first->second != t[i]) return false; } auto it2 = mp2.insert(make_pair(t[i], s[i])); if(!it2.second){ if(it2.first->second != s[i]) return false; } } return true;*/ };

解法二：使用unordered_map，时间复杂度O(n)：

class Solution { public: bool isIsomorphic(string s, string t) { if(s.length() != t.length()) return false; unordered_map<char, char> umap1; unordered_map<char, char> umap2; for(int i=0; i<s.length(); ++i){ auto it1 = umap1.insert(make_pair(s[i], t[i])); if(!it1.second){ if(it1.first->second != t[i]) return false; } auto it2 = umap2.insert(make_pair(t[i], s[i])); if(!it2.second){ if(it2.first->second != s[i]) return false; } } };

别人的解法：)，也是O(n)：

class Solution { public: bool isIsomorphic(string s, string t) { int m1[256] = {0}, m2[256] = {0}; for(int i=0; i<s.length(); ++i) { if(m1[s[i]] != m2[t[i]]) return false; m1[s[i]] = i + 1; m2[t[i]] = i + 1; } return true; } };