Description of the 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, 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.

The idea of solving the problem:

It is not difficult to see that:

1. When the current number is smaller than the right one, it subtracts the current number.

2. When the current number is larger than the right digit or equal to the right number, add the current number.

And 12. Integer to Roman can form a series

Code:

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