Leetcode -- integer to Roman

Given an integer, convert it to a roman numeral.

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

Given an integer, convert it to a Roman number.

The input value can be between 1 and 3999.

Is a conversion rule.

1	2	3	4	5	6	7	8	9
I	II	III	IV	V	VI	VII	VIII	IX
10	20	30	40	50	60	70	80	90
X	Xx	Xxx	XL	L	Lx	LXX	Lxxx	XC
100	200	300	400	500	600	700	800	900
C	CC	CCC	CD	D	DC	DCC	DCCC	Cm

The following program uses the example code to write the above table using if else. Of course, this course schedule is also regular. For example, the course starts with 1-3 and repeats. The last character is intercepted from 4-5, and the last character is added from 6 to 8.

public static String intToRoman(int num) {StringBuffer buf = new StringBuffer();int dddd = num / 1000;int ddd = num % 1000 / 100;int dd = num % 1000 % 100 / 10;int d = num % 1000 % 100 % 10;for(int i=0;i<dddd;i++)buf.append("M");if(ddd == 1)buf.append("C");else if(ddd == 2)buf.append("CC");else if(ddd == 3)buf.append("CCC");else if(ddd == 4)buf.append("CD");else if(ddd == 5)buf.append("D");else if(ddd == 6)buf.append("DC");else if(ddd == 7)buf.append("DCC");else if(ddd == 8)buf.append("DCCC");else if(ddd == 9)buf.append("CM");if(dd == 1)buf.append("X");else if(dd == 2)buf.append("XX");else if(dd == 3)buf.append("XXX");else if(dd == 4)buf.append("XL");else if(dd == 5)buf.append("L");else if(dd == 6)buf.append("LX");else if(dd == 7)buf.append("LXX");else if(dd == 8)buf.append("LXXX");else if(dd == 9)buf.append("XC");if(d == 1)buf.append("I");else if(d == 2)buf.append("II");else if(d == 3)buf.append("III");else if(d == 4)buf.append("IV");else if(d == 5)buf.append("V");else if(d == 6)buf.append("VI");else if(d == 7)buf.append("VII");else if(d == 8)buf.append("VIII");else if(d == 9)buf.append("IX");return buf.toString();}

See this http://www.mathsisfun.com/roman-numerals.html for conversion rules for integers and Roman numerals.