整数转罗马数字

七个不同的符号代表罗马数字，其值如下：

符号	值
I	1
V	5
X	10
L	50
C	100
D	500
M	1000

罗马数字是通过添加从最高到最低的小数位值的转换而形成的。将小数位值转换为罗马数字有以下规则：

如果该值不是以 4 或 9 开头，请选择可以从输入中减去的最大值的符号，将该符号附加到结果，减去其值，然后将其余部分转换为罗马数字。 如果该值以 4 或 9 开头，使用 减法形式，表示从以下符号中减去一个符号，例如 4 是 5 (V) 减 1 (I): IV ，9 是 10 (X) 减 1 (I)：IX。仅使用以下减法形式：4 (IV)，9 (IX)，40 (XL)，90 (XC)，400 (CD) 和 900 (CM)。 只有 10 的次方（I, X, C, M）最多可以连续附加 3 次以代表 10 的倍数。你不能多次附加 5 (V)，50 (L) 或 500 (D)。如果需要将符号附加4次，请使用 减法形式。 给定一个整数，将其转换为罗马数字。

示例 1：

输入：num = 3749 输出： "MMMDCCXLIX" 解释： 3000 = MMM 由于 1000 (M) + 1000 (M) + 1000 (M) 700 = DCC 由于 500 (D) + 100 (C) + 100 (C) 40 = XL 由于 50 (L) 减 10 (X) 9 = IX 由于 10 (X) 减 1 (I) 注意：49 不是 50 (L) 减 1 (I) 因为转换是基于小数位

示例 2：

输入：num = 58 输出："LVIII" 解释： 50 = L 8 = VIII

示例 3：

输入：num = 1994 输出："MCMXCIV" 解释： 1000 = M 900 = CM 90 = XC 4 = IV

提示： 1 <= num <= 3999

解题思路：

将每个数位对应的罗马字符串预存为二维表，table[位数][数字]，位数 0=个位，1=十位，2=百位，3=千位，按位提取数字直接查表拼接。

C++

class Solution {
public:
    string intToRoman(int num) {
        const string table[4][10] = {
            {"","I","II","III","IV","V","VI","VII","VIII","IX"},
            {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
            {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
            {"","M","MM","MMM","","","","","",""},
        };
        string s = to_string(num);
        int n = s.size();
        string res;
        for (int i = 0; i < n; i++) {
            int position = n - 1 - i;
            int digit = s[i] - '0';
            res += table[position][digit];
        }
        return res;
    }
};

Python

def intToRoman(num: int) -> str:
    table: list[list[str]] = [
        ["","I","II","III","IV","V","VI","VII","VIII","IX"],
        ["","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"],
        ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"],
        ["","M","MM","MMM"],
    ]
    s: str = str(num)
    n: int = len(s)
    res: str = ""
    for i in range(n):
        position: int = n - 1 - i
        digit: int = int(s[i])
        res += table[position][digit]
    return res

Go

func intToRoman(num int) string {
    var table [][]string = [][]string{
        {"","I","II","III","IV","V","VI","VII","VIII","IX"},
        {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
        {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
        {"","M","MM","MMM"},
    }
    s := strconv.Itoa(num)
    n := len(s)
    res := ""
    for i := 0; i < n; i++ {
        position := n - 1 - i
        digit := int(s[i] - '0')
        res += table[position][digit]
    }
    return res
}

Java

class Solution {
    public String intToRoman(int num) {
        String[][] table = {
            {"","I","II","III","IV","V","VI","VII","VIII","IX"},
            {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
            {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
            {"","M","MM","MMM"},
        };
        String s = String.valueOf(num);
        int n = s.length();
        String res = "";
        for (int i = 0; i < n; i++) {
            int position = n - 1 - i;
            int digit = s.charAt(i) - '0';
            res += table[position][digit];
        }
        return res;
    }
}

时间 O(1)（数字范围固定），空间 O(1)。