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LeetCode--Roman to Integer

Roman to Integer

1. 題目

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. 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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

2. 分析

這題實現起來不算難,只要注意一些規律,就是當前一個數字比後一個數字要小的時候,那麼這兩個數字必須是一個組合。然後map這個結構完美契合這個題目。

3. 原始碼

class Solution {
public:
    int romanToInt(string s) {
        map<char, int> m;
        int sum = 0;
        m['I'] = 1;
        m['V'] = 5;
        m['X'] = 10;
  		m['L'] = 50;
        m['C'] = 100;
        m['D'] = 500;
        m['M'] = 1000;
        for(int i = 0; i < s.size(); i++) {
        	int next = i+1;
        	if(next < s.size()) {
        		if(m[s[next]] > m[s[i]]) {
        			sum += (m[s[next]] - m[s[i]]);
        			i++;
        		} else {
        			sum +=  m[s[i]];
        		}
        	} else {
        		sum += m[s[i]];
        	}
        }
        return sum;
    }
};