leetcode 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.

我的解决方案：

// isIsomorphic.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"


#include<map>
#include<string>
#include<iostream>
#include<unordered_map>

using namespace std;



bool isIsomorphic(string s, string t) 
{
    if(s.length()!=t.length())return false;

    int s_length = s.length();
    int t_length = t.length();
    unordered_map<char,char> stemp;
    unordered_map<char,char> ttemp;

    for(int i = 0;i <  s_length; i++)
    {
        if(stemp.find(s[i]) == stemp.end() && ttemp.find(t[i]) == ttemp.end())
        {
            stemp[s[i]] = t[i];
            ttemp[t[i]] = s[i];
        }
        else
        {
            if(stemp.find(s[i]) == stemp.end() && ttemp[t[i]]!=s[i])
            { 
                return false; 
            }
            else if(ttemp.find(t[i])==ttemp.end() && stemp[s[i]]!=t[i])
            {
                return false; 
            }
            else if(stemp[s[i]] != t[i] && ttemp[t[i]] != s[i])
            { 
                return false; 
            }

        }
    }
}
//
//pair<map<char,int>::iterator,bool> Insert_Pair;
//Insert_Pair = mapString.insert(map<char,int>::value_type(s[i],(int)(s[i] - t[i])));


int _tmain(int argc, _TCHAR* argv[])
{

    string s = "ab";
    string t = "aa";

    isIsomorphic(s,t);
    return 0;
}

unordered_map 简介：
http://blog.csdn.net/gamecreating/article/details/7698719
http://blog.csdn.net/orzlzro/article/details/7099231
http://blog.csdn.net/sws9999/article/details/3081478

unordered_map，它与map的区别就是map是按照operator<比较判断元素是否相同，以及比较元素的大小，然后选择合适的位置插入到树中。所以，如果对map进行遍历（中序遍历）的话，输出的结果是有序的。顺序就是按照operator< 定义的大小排序。而unordered_map是计算元素的Hash值，根据Hash值判断元素是否相同。所以，对unordered_map进行遍历，结果是无序的。而hash则是把数据的存储和查找消耗的时间大大降低；而代价仅仅是消耗比较多的内存。虽然在当前可利用内存越来越多的情况下，用空间换时间的做法是值得的。
用法的区别就是map的key需要定义operator<。而unordered_map需要定义hash_value函数并且重载operator==。对于自定义的类型做key，就需要自己重载operator< 或者hash_value()了。

python 的解决方案：

def isIsomorphic(self, s, t):
    if len(s) != len(t):
        return False
    def halfIsom(s, t):
        res = {}
        for i in xrange(len(s)):
            if s[i] not in res:
                res[s[i]] = t[i]
            elif res[s[i]] != t[i]:
                return False
        return True
    return halfIsom(s, t) and halfIsom(t, s)