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)