No.205 Ismorphic Strings

No.205 Ismorphic 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.

　　字符串同构判断
　　已知s和t大小相同，s中的字符经过相同的映射可转换为t，则s和t是同构的
　　注：两个字符不能映射到同一字符，但字符可映射到它本身

#include "stdafx.h"
#include <map>
#include <vector>
#include <iostream>
using namespace std;

class Solution
{
public:
    bool isIsomorphic(string s, string t)
    {//字符串同构判断
     //已知s和t大小相同，s中的字符经过相同的映射可转换为t，则s和t是同构的
     //注：两个字符不能映射到同一字符，但字符可映射到它本身
     int size = s.size();
     if(size <= 1)
         return true;//大小为1，肯定同构
     map<char,char> mapping;
     for(int i=0; i<size; i++)
     {
         if(mapping.find(s[i]) == mapping.end())//尚未建立映射
         {
             for(auto const &it : mapping)
                if(it.second == t[i])//两个字符映射到同一字符了!!
                    return false;
             mapping[s[i]] = t[i];
         }
         else
             if(t[i] != mapping[s[i]])
                 return false;
         
     }
     return true;
    }
};

int main()
{
    Solution sol;
    cout << boolalpha;

    cout << sol.isIsomorphic("egg","add")<<endl;//true
    cout << sol.isIsomorphic("foo","bar")<<endl;//false
    cout << sol.isIsomorphic("paper","title")<<endl;//true
    cout << sol.isIsomorphic("abcda","bcceb")<<endl;//两个字符映射相同
}