12、Integer to Roman

题目：

Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

解题思路：

罗马数字共有七个，即I(1)，V(5)，X(10)，L(50)，C(100)，D(500)，M(1000)。按照下面的规则可以表示任意正整数。重复数次：一个罗马数字重复几次，就表示这个数的几倍。 右加左减：在一个较大的罗马数字的右边记上一个较小的罗马数字，表示大数字加小数字。在一个较大的数字的左边记上一个较小的罗马数字，表示大数字减小数字。但是，左减不能跨越等级。比如，99不可以用IC表示，用XCIX表示。 加线乘千：在一个罗马数字的上方加上一条横线或者在右下方写M，表示将这个数字乘以1000，即是原数的1000倍。同理，如果上方有两条横线，即是原数的1000000倍。 单位限制：同样单位只能出现3次，如40不能表示为XXXX，而要表示为XL。

python版本：

class Solution(object):    def intToRoman(self, num):        """        :type num: int        :rtype: str        """        roman = [            ["","I","II","III","IV","V","VI","VII","VIII","IX"],            ["","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"],            ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"],            ["","M","MM","MMM"]            ]        digit = 0        res = ""        while num>0:            remain = num%10            res = roman[digit][remain]+res            num = num/10            digit += 1        return res

c++版本：

class Solution {public:    string intToRoman(int num) {        string roman[][10] = {            {"","I","II","III","IV","V","VI","VII","VIII","IX"},            {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},            {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},            {"","M","MM","MMM"}            };                                      //1-9,10-90,100-900,1000-3000        string res = "";        int digit = 0;        while(num>0)        {            int remain = num%10;            res = roman[digit][remain]+res;            digit++;            num = num/10;        }        return res;    }};

罗
马
数字共有七个，即I(1)，V(5)，X(10)，L(50)，C(100)，D(500)，M(1000)。按照下面的规则可以表示任意正整数。重复数次：一个罗马数字重复几次，就表示这个数的几倍。 右加左减：在一个较大的罗马数字的右边记上一个较小的罗马数字，表示大数字加小数字。在一个较大的数字的左边记上一个较小的罗马数字，表示大数字减小数字。但是，左减不能跨越等级。比如，99不可以用IC表示，用XCIX表示。 加线乘千：在一个罗马数字的上方加上一条横线或者在右下方写M，表示将这个数字乘以1000，即是原数的1000倍。同理，如果上方有两条横线，即是原数的1000000倍。 单位限制：同样单位只能出现3次，如40不能表示为XXXX，而要表示为XL。