[leetcode] Roman to Integer (c++)

Problem

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 one's 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

 

Example 2:

Input: s = "IV"

Output: 4

 

Example 3:

Input: s = "IX"

Output: 9

 

Example 4:

Input: s = "LVIII"

Output: 58

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

 

Example 5:

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].

How to solve

로만 symbol로 나타낸 글자를 Integer값으로 변환하는 문제!

* 4, 9의 표기에 유의해야한다. 

 

1. map에 symbol을 저장한다. 

2. 결과 값을 담을 변수(res)를 선언하고, 마지막 symbol에 대한 값을 넣는다. 

2. String을 순회하며 i 번째 숫자보다 i+1번째 숫자가 더 크면 res에 현재 값을 +, 작으면 -를 한다. 

 

Solution(c++)

#include <bits/stdc++.h>

class Solution {
public:
    int romanToInt(string s) {
        unordered_map<char, int> RtoI ={{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
        int res = RtoI[s.back()];
        
        for(int i=0; i<s.length()-1; i++){
            RtoI[s[i]]>=RtoI[s[i+1]]? res += RtoI[s[i]] : res -= RtoI[s[i]];
        }
        return res;
    }
};

Test Result