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.
Solution1: Simulation
Let's walk through an example, if input: 6, output is "LI", we find the maximum value that we can remove from 6, iteratively doing that and appending its corresponding Roman numerals into result.
Step1, For most cases, Roman Numeral use addition(相加).
Coz only a few cases using subtraction(相减) like 4 or 9, we can pre-calculate those ones.
int values[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
String romans[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
Step2, traverse the values[], find the maximum value we can remove from 6
int values[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
String romans[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
Step3, append its corresponding Roman numerals into result, remove it from 6, and jump to Step2 to do the next iteration
code:
1 /* 2 Time Complexity: O(n) 3 Space Complexity: O(1) 4 */ 5 class Solution { 6 public String intToRoman(int num) { 7 final int values[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}; 8 final String romans[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}; 9 StringBuilder sb = new StringBuilder(); 10 for (int i = 0; i < values.length; i++){ 11 while( num >= values[i]){ 12 num -= values[i]; 13 sb.append(romans[i]); 14 } 15 } 16 return sb.toString(); 17 } 18 }
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