Description#
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, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 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 before V (5) and X (10) to make 4 and 9. X can be placed before L (50) and C (100) to make 40 and 90. C can be placed before D (500) and M (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.
Example 1:
Input: num = 3
Output: "III"
Explanation: 3 is represented as 3 ones.
Example 2:
Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 3:
Input: num = 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
Constraints:
Solutions#
Solution 1: Greedy#
We can first list all possible symbols $cs$ and their corresponding values $vs$, then enumerate each value $vs[i]$ from large to small. Each time, we use as many symbols $cs[i]$ corresponding to this value as possible, until the number $num$ becomes $0$.
The time complexity is $O(m)$, and the space complexity is $O(m)$. Here, $m$ is the number of symbols.
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| class Solution:
def intToRoman(self, num: int) -> str:
cs = ('M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I')
vs = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1)
ans = []
for c, v in zip(cs, vs):
while num >= v:
num -= v
ans.append(c)
return ''.join(ans)
|
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| class Solution {
public String intToRoman(int num) {
List<String> cs
= List.of("M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I");
List<Integer> vs = List.of(1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1);
StringBuilder ans = new StringBuilder();
for (int i = 0, n = cs.size(); i < n; ++i) {
while (num >= vs.get(i)) {
num -= vs.get(i);
ans.append(cs.get(i));
}
}
return ans.toString();
}
}
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| class Solution {
public:
string intToRoman(int num) {
vector<string> cs = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
vector<int> vs = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
string ans;
for (int i = 0; i < cs.size(); ++i) {
while (num >= vs[i]) {
num -= vs[i];
ans += cs[i];
}
}
return ans;
}
};
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| func intToRoman(num int) string {
cs := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
vs := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
ans := &strings.Builder{}
for i, v := range vs {
for num >= v {
num -= v
ans.WriteString(cs[i])
}
}
return ans.String()
}
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| function intToRoman(num: number): string {
const cs: string[] = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
const vs: number[] = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
const ans: string[] = [];
for (let i = 0; i < vs.length; ++i) {
while (num >= vs[i]) {
num -= vs[i];
ans.push(cs[i]);
}
}
return ans.join('');
}
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| public class Solution {
public string IntToRoman(int num) {
List<string> cs = new List<string>{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
List<int> vs = new List<int>{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
StringBuilder ans = new StringBuilder();
for (int i = 0; i < cs.Count; i++) {
while (num >= vs[i]) {
ans.Append(cs[i]);
num -= vs[i];
}
}
return ans.ToString();
}
}
|
