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13. 罗马数字转整数


13. 罗马数字转整数

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题目

Roman numerals are represented by seven different symbols: IVXLCD 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 a roman numeral, convert it to an integer.

Example 1:

Input: s = "III"

Output: 3

Explanation: III = 3.

Example 2:

Input: s = "LVIII"

Output: 58

Explanation: L = 50, V= 5, III = 3.

Example 3:

Input: s = "MCMXCIV"

Output: 1994

Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Constraints:

  • 1 <= s.length <= 15
  • s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
  • It is guaranteed that s is a valid roman numeral in the range [1, 3999].

题目大意

罗马数字包含以下七种字符: I, V, X, L, C, D, M

通常情况下,罗马数字中小的数字在大的数字的右边。但也存在特例:

  • I  可以放在  V (5) 和  X (10) 的左边,来表示 49
  • X  可以放在  L (50) 和  C (100) 的左边,来表示 40 90
  • C  可以放在  D (500) 和  M (1000) 的左边,来表示 400900

给定一个罗马数字,将其转换成整数。输入确保在 1  到 3999 的范围内。

解题思路

按照题目中罗马数字的字符数值,计算出对应罗马数字的十进制数即可。

代码

/**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function (s) {
	const roman = {
		I: 1,
		V: 5,
		X: 10,
		L: 50,
		C: 100,
		D: 500,
		M: 1000
	};

	let num, nextNum;
	let total = 0;

	for (let i = 0; i < s.length; i++) {
		num = roman[s[i]];
		nextNum = roman[s[i + 1]];
		if (num < nextNum) {
			total -= num;
		} else {
			total += num;
		}
	}

	return total;
};

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