0012: Integer to Roman
Problem Statement
Seven different symbols represent Roman numerals with the following values:
Symbol | Value |
---|---|
I | 1 |
V | 5 |
X | 10 |
L | 50 |
C | 100 |
D | 500 |
M | 1000 |
Roman numerals are formed by appending the conversions of decimal place values from highest to lowest. Converting a decimal place value into a Roman numeral has the following rules:
- If the value does not start with 4 or 9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.
- If the value starts with 4 or 9 use the subtractive form representing one symbol subtracted from the following symbol, for example, 4 is 1 (
I
) less than 5 (V
):IV
and 9 is 1 (I
) less than 10 (X
):IX
. Only the following subtractive forms are used: 4 (IV
), 9 (IX
), 40 (XL
), 90 (XC
), 400 (CD
) and 900 (CM
). - Only powers of 10 (
I
,X
,C
,M
) can be appended consecutively at most 3 times to represent multiples of 10. You cannot append 5 (V
), 50 (L
), or 500 (D
) multiple times. If you need to append a symbol 4 times use the subtractive form.
Given an integer, convert it to a Roman numeral.
Example 1:
Input: num = 3749
Output: "MMMDCCXLIX"
Explanation:
3000 = MMM as 1000 (M) + 1000 (M) + 1000 (M) 700 = DCC as 500 (D) + 100 (C) + 100 (C) 40 = XL as 10 (X) less of 50 (L) 9 = IX as 1 (I) less of 10 (X) Note: 49 is not 1 (I) less of 50 (L) because the conversion is based on decimal places
Example 2:
Input: num = 58
Output: "LVIII"
Explanation:
50 = L 8 = VIII
Example 3:
Input: num = 1994
Output: "MCMXCIV"
Explanation:
1000 = M 900 = CM 90 = XC 4 = IV
Constraints:
1 <= num <= 3999
Code Solution
class Solution:
def intToRoman(self, num):
keys = [
1000, 900, 500, 400, 100,
90, 50, 40, 10, 9, 5, 4, 1
]
vals = [
'M', 'CM', 'D', 'CD', 'C',
'XC', 'L', 'XL', 'X', 'IX',
'V', 'IV', 'I'
]
cache = dict(zip(keys, vals))
ix = 0
out = []
while num:
while num < keys[ix]:
ix = ix + 1
num = num - keys[ix]
out.append(vals[ix])
return "".join(out)