LeetCode | 12. Integer to Roman

最新推荐文章于 2022-03-25 19:34:53 发布

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最新推荐文章于 2022-03-25 19:34:53 发布

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本文详细介绍了一种将整数转换为罗马数字的算法实现，通过分析罗马数字的特殊构成规则，如IV代表4、IX代表9等，展示了如何使用C++编程语言进行高效转换。文章通过具体实例解释了算法的工作原理，包括从千位到个位的逐级转换过程。

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题目：

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 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. 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.

代码：

class Solution {
public:
    string intToRoman(int num) {
        string res = "";
		int base[] = { 1000, 500, 100, 50, 10, 5, 1 };
		char r[] = { 'M', 'D', 'C', 'L', 'X', 'V', 'I' };
		for (int i = 0; i < 7; i += 2)
		{
			int q = num / base[i];
			if (q == 9)
			{
				res += r[i];
				res += r[i - 2];
				q -= 9;
			}
			else if (q > 4)
			{
				res += r[i - 1];
				q -= 5;
			}
			else if (q == 4)
			{
				res += r[i];
				res += r[i - 1];
				q -= 4;
			}

			while (q--)
				res += r[i];
			num %= base[i];
		}
		return res;
    }
};