【leetcode刷题笔记】Integer to Roman

Given an integer, convert it to a roman numeral.

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

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题解：基本的罗马字符和数字对应如下表所示：

罗马字符	数字
I	1
V	5
X	10
L	50
C	100
D	500
M	1000

每隔一个字符又可以与相邻的两个字符组成一个新的数，例如I可以和V和X组成IV和IX，这样又可以生成6个数字，如下表：

罗马字符	数字
IV	4
IX	9
XL	40
XC	90
CD	400
CM	900

接下来我们用numbers记录上述13个数字，symbol数组记录上述13个符号，然后从最大的数1000开始，一步步将num转换成数字。

数字3012的转换过程如下：

1. 3012/1000 = 3，answer = MMM，num <- 3012-3000=12;
2. 12/900=0; 12/500 = 0; ......
3. 12/10 = 1, answer = MMM+X=MMMX, num <- 12 - 10 = 2;
4. 2/9 = 0,; 2/5 = 0; 2/4 = 0;
5. 2/1 = 2, answer = MMMX+II = MMMXII, num <- 2-2 = 0;

所以3012对应的罗马数字为MMMXII。

代码如下：

 1 public class Solution {

 2     public String intToRoman(int num) {

 3         if(num <= 0)

 4             return "";

 5         

 6         int[] numbers = {1000,900,500,400,100,90,50,40,10,9,5,4,1};

 7         String[] symbol = {"M","CM","D","CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};

 8         

 9         StringBuffer answer = new StringBuffer();

10         int digit = 0;

11         while(num > 0){

12             int times = num / numbers[digit];

13             for(int i = 0;i < times;i++)

14                 answer.append(symbol[digit]);

15             num = num - numbers[digit] * times;

16             digit++;

17         }

18         return answer.toString();

19     }

20 }