标签：leetcode
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
给定一个整数，把它转换成罗马数字。
输入可以保证在1到3999之间。
下图是转换规则。
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
|
I
|
II
|
III
|
IV
|
V
|
VI
|
VII
|
VIII
|
IX
|
|
10
|
20
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30
|
40
|
50
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60
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70
|
80
|
90
|
|
X
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XX
|
XXX
|
XL
|
L
|
LX
|
LXX
|
LXXX
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XC
|
|
100
|
200
|
300
|
400
|
500
|
600
|
700
|
800
|
900
|
|
C
|
CC
|
CCC
|
CD
|
D
|
DC
|
DCC
|
DCCC
|
CM
|
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();
}LeetCode——Integer to Roman,布布扣,bubuko.com
标签：leetcode
原文地址：http://blog.csdn.net/laozhaokun/article/details/35579687
