205. Isomorphic Strings

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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;    }};

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