Roman numerals are represented by seven different symbols: I
, V
, X
, L
, C
, D
and M
.
Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000
For example, two is written as II
in Roman numeral, just two one's added together. Twelve is written as, XII
, which is simply X
+ II
. The number twenty seven is written as XXVII
, which is XX
+ V
+ II
.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII
. Instead, the number four is written as IV
. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX
. There are six instances where subtraction is used:
-
I
can be placed beforeV
(5) andX
(10) to make 4 and 9. -
X
can be placed beforeL
(50) andC
(100) to make 40 and 90. -
C
can be placed beforeD
(500) andM
(1000) to make 400 and 900.
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
Example 1:
Input: 3 Output: "III"
Example 2:
Input: 4 Output: "IV"
Example 3:
Input: 9 Output: "IX"
Example 4:
Input: 58 Output: "LVIII" Explanation: L = 50, V = 5, III = 3.
Example 5:
Input: 1994 Output: "MCMXCIV" Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
思路:
整形转罗马数字时,1,2,..,9,10,20,...,90,100,200,...,900,...的转换方式是相同的,不同的只是对应的字母。因此将一个整数转换为罗马数字只需将整数分开成个位、十位...单独转换,最后在整合在一起即可。
代码:
//Integer to Roman public static String intToRoman(int num){ List<String> l = new ArrayList<>(); int flag = 0; String a = "X",b = "V",c = "I"; while (num!=0) { int n = num%10; num = num/10; if(flag == 1){ a = "C";b="L";c="X"; } if(flag == 2){ a = "M";b="D";c="C"; } if(flag == 3){ c = "M"; } String str = romanSub(n,a,b,c); l.add(str); flag ++; } String result = ""; for (int i = l.size()-1; i >= 0; i--) { result+=l.get(i); } return result; } //a 10 b 5 c 1 public static String romanSub(int num,String a,String b,String c){ String res = ""; if(num<5){ if(num<4){ for (int i = 0; i < num; i++) { res += c; } }else{ res = c+b; } }else if(num == 5){ res = b; }else{ if(num<9){ res = b; for (int i = 5; i < num; i++) { res += c; } }else{ res = c+a; } } return res; }