LeetCode算法題之int to Roman

問題描述：
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
將數字轉換爲羅馬數字表示。
解題思路：
按照羅馬數字組數標準組合即可，附羅馬數字百度百科：http://baike.baidu.com/link?url=UUF_EffApqqZ3Hh2K_7f8NOYpFYIAtk8oWQMWkmN1KnO3TOeiitCwubqKI_boYSlsKOEij4zrwMT0oGN4Ev0YK

class Solution
{
public:

    string intToRoman(int num)
    {
        map<int, char> Int_Roman;
        Int_Roman.insert(pair<int,char>(1,'I'));
        Int_Roman.insert(pair<int,char>(5,'V'));
        Int_Roman.insert(pair<int,char>(10,'X'));
        Int_Roman.insert(pair<int,char>(50,'L'));
        Int_Roman.insert(pair<int,char>(100,'C'));
        Int_Roman.insert(pair<int,char>(500,'D'));
        Int_Roman.insert(pair<int,char>(1000,'M'));
        string result;
        if(num<0 || num >3999)
            return "";
        //確定個十百千位，他們相加起來正好等於num
        int single = num % 10;
        int ten = num % 100 - single;
        int hundred = num % 1000 - num % 100;
        int thousand = num - num % 100;
        //確定千位
        for(int j=0; j<thousand/1000; j++ )
        {
            result += Int_Roman[1000];
        }
        //確定百位
        if(hundred == 900)
        {
            result += "CM";
        }
        else if(hundred == 400)
        {
            result += "CD";

        }
        else if(hundred < 400)
        {
            for(int j=0; j<hundred/100; j++ )
            {
                result += Int_Roman[100];
            }
        }
        else
        {
            result += 'D';
            for(int j=0; j<(hundred - 500)/100; j++ )
            {
                result += Int_Roman[100];
            }
        }
        //確定十位
        if(ten == 90)
        {
            result += "XC";
        }
        else if(ten == 40)
        {
            result += "XL";

        }
        else if(ten < 40)
        {
            for(int j=0; j<ten/10; j++ )
            {
                result += Int_Roman[10];
            }
        }
        else
        {
            result += 'L';
            for(int j=0; j<(ten - 50)/10; j++ )
            {
                result += Int_Roman[10];
            }
        }
        //確定個位
        if(single == 9)
        {
            result += "IX";
        }
        else if(single == 4)
        {
            result += "IV";

        }
        else if(single < 4)
        {
            for(int j=0; j<single/1; j++ )
            {
                result += Int_Roman[1];
            }
        }
        else
        {
            result += 'V';
            for(int j=0; j<(single - 5)/1; j++ )
            {
                result += Int_Roman[1];
            }
        }
//.....................................................
    return result;
    }
};