leetcode : 13. Roman to Integer

最新推荐文章于 2025-11-30 14:38:30 发布

原创最新推荐文章于 2025-11-30 14:38:30 发布 · 184 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种将罗马数字转换为整数的方法，通过使用HashMap数据结构存储罗马数字与整数值的对应关系，实现了准确的转换。文章详细解释了解题思路，包括如何处理特殊规则，如IV表示4而不是IIII，提供了完整的Java代码实现。

题目标签

string

题目难度

easy

题目地址：

https://leetcode.com/problems/roman-to-integer/

题目描述 :

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 a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:
Input: "III"
Output: 3

Example 2:
Input: "IV"
Output: 4

Example 3:
Input: "IX"
Output: 9

Example 4:
Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:
Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

解题思路

因题中是罗马数字与阿拉伯数字值一一对应，是键值对的关系，所以我打算采用HashMap的数据结构，将其进行存储。
正如题中所说罗马数字正常是从左向右依次减小的，所以从HashMap中取出每个字符对应的值，并且依次相加。
因为一些特例，而根据第2步，会出现一些错误，如：MCM ->1000+100+1000；实际应该为1000+900；但是，第二步已经将C的值加上了，所以这一步我将如此处理：1000+100-100*2+1000; 这个意思是：当碰到下一字符所对应的值比上一字符所对应的值大的时候，将上一个字符对应的值双倍减掉。
完成。

代码

Java

class Solution {
    public int romanToInt(String s) {
        if (s == null || "".equals(s.trim())) {
            return -1;
        }
        HashMap<Character, Integer> map = new HashMap<>();
        map.put('I', 1);
        map.put('V', 5);
        map.put('X', 10);
        map.put('L', 50);
        map.put('C', 100);
        map.put('D', 500);
        map.put('M', 1000);
        int result = 0;
        int tmp = 0;
        int number = 0;
        for (int i = 0; i < s.length(); i++) {
            tmp = map.get(s.charAt(i));
            if (tmp > number && number != 0) {
                result -= 2 * number;
            }
            result += tmp;
            number = tmp;
        }
        return result;
    }
}