13. Roman to Integer

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本文介绍了一种将罗马数字转换为整数的方法。通过使用哈希表存储罗马数字及其对应的值，结合特定规则进行加减运算，实现了从罗马数字到整数的有效转换。例如，IV表示4，MCMXCIV表示1994。

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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.

解题思路：将罗马数字转换成数字，小在左边是减，小的在右边是加。用unordered_map存储键值对，存储罗马数字代表多少

class Solution {
public:
    int romanToInt(string s) {
     unordered_map<char,int>k;
     k.insert(make_pair('I',1));
     k.insert(make_pair('V',5));
     k.insert(make_pair('X',10));
     k.insert(make_pair('L',50));
     k.insert(make_pair('C',100));
     k.insert(make_pair('D',500));
     k.insert(make_pair('M',1000));
     int count=0;
     int i=0;
        while(i<s.size()){
            if(k[char(s[i])]<k[char(s[i+1])]){
                count+=k[char(s[i+1])]-k[char(s[i])];
                i+=2;
            }
            else{
                count+=k[char(s[i])];
                i++;
            }
        }
     return count;
         
    
    }
};