整数转罗马数字

Posted on May 21, 2026 By ᵇᵒ

Integer to Roman

LeetCode 12

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

Approach 1

class Solution {
    private val symbols = arrayOf(
        1000 to "M", 900 to "CM", 500 to "D", 400 to "CD",
        100 to "C", 90 to "XC", 50 to "L", 40 to "XL",
        10 to "X", 9 to "IX", 5 to "V", 4 to "IV",
        1 to "I"
    )

    fun intToRoman(num: Int): String = buildString {
        symbols.fold(num) { acc, (n, str) ->
            val times = acc / n
            repeat(times) { append(str) }
            acc - n * times
        }
    }
}

Approach 2

class Solution {
    private val m = arrayOf("", "M", "MM", "MMM")
    private val c = arrayOf("", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM")
    private val x = arrayOf("", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC")
    private val i = arrayOf("", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX")

    fun intToRoman(n: Int): String {
        return m[n / 1000] + c[(n % 1000) / 100] + x[(n % 100) / 10] + i[n % 10]
    }
}

或者更统一的写法：

class Solution {
    private val m = arrayOf("", "M", "MM", "MMM")
    private val c = arrayOf("", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM")
    private val x = arrayOf("", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC")
    private val i = arrayOf("", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX")

    private val triples = arrayOf(
        Triple(m, 4000, 1000),
        Triple(c, 1000, 100),
        Triple(x, 100, 10),
        Triple(i, 10, 1),
    )

    fun intToRoman(n: Int): String = buildString {
        triples.forEach { (arr, m, d) -> append(arr[(n % m) / d]) }
    }
}