[Leetcode] Isomorphic Strings

题目链接在此

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.

第一个想到的肯定是用哈希，就是用map逐个做映射，遇到映射错误或多对一的情况就return false：

class Solution {public:bool isIsomorphic(string s, string t) {if (s.size() != t.size())return false;map<char, char> m;set<char> mapped;for (int i = 0; i < s.size(); i++) {if (m.find(s[i]) != m.end() && m[s[i]] == t[i])continue;if (m.find(s[i]) == m.end() && mapped.find(t[i]) == mapped.end()) {m[s[i]] = t[i];mapped.insert(t[i]);}elsereturn false;}return true;}};

但这样还是有点慢。于是找到了这位大神的解法——把相同的"pattern"映射成一个新"pattern"，比如把"egg"和"add"都映射成"122"。

大神说不得转载，那咱就只好自己写一份了。

class Solution {public:bool isIsomorphic(string s, string t) {if (s.size() != t.size())return false;return convert(s) == convert(t);}private:string convert(string str) {char c = '1';char mapping[128] = { 0 };for (int i = 0; i < str.size(); i++) {if (mapping[str[i]] == 0) {mapping[str[i]] = c++;}str[i] = mapping[str[i]];}return str;}};