Leetcode 13. Roman to Integer

最新推荐文章于 2025-12-26 21:45:18 发布

原创最新推荐文章于 2025-12-26 21:45:18 发布 · 320 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #算法

algorithm-design 专栏收录该内容

817 篇文章

订阅专栏

本文介绍如何将罗马数字（如III、LVIII、MCMXCIV）转换为整数，通过举例和代码实现了一个Solution类，展示了罗马数字中减法规则的应用。

Roman to Integer
Easy
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 ones 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 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].

解法1：

class Solution {
public:
    int romanToInt(string s) {
        int n = s.size();
        int sum = 0;
        char c1 = 'Z', c2 = 'Z';
        for (int i = 0; i < n; i++) {
            c1 = s[i];
            if (i < n - 1) c2 = s[i + 1];
            switch(c1) {
                case 'I':
                    if (c2 == 'V' || c2 == 'X') sum -= 1;
                    else sum += 1;
                    break;
                case 'V':
                    sum += 5;
                    break;
                case 'X':
                    if (c2 == 'L' || c2 == 'C') sum -= 10;
                    else sum += 10;
                    break;
                case 'L':
                    sum += 50;
                    break;
                case 'C':
                    if (c2 == 'D' || c2 == 'M') sum -= 100;
                    else sum += 100;
                    break;
                case 'D':
                    sum += 500;
                    break;
                case 'M':
                    sum += 1000;
                    break;
                default:
                    break;
            }
        }
        return sum;
    }
};