LeetCode 13 Roman to Integer (Rome to Integer)

LeetCode 13 Roman to Integer (Rome to Integer)

Translation

Given a roman number, convert it to an integer value. The input value must be between 1 and 3999.

Original

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

I didn't have a good idea at the beginning. Although I can continue using the previous topic, the result is as follows ...... Messy code ......

public class Solution{    public int RomanToInt(string s)    {        int result = 0;        Type R = typeof(Roman);        string first, second;        if (s.Length > 1)        {            first = s.Substring(0, 1); second = s.Substring(0, 2);        }        else        {            first = s.Substring(0, 1); second = ;        }        foreach (var r in Enum.GetNames(R).Reverse())        {            while ((r.Length == 1 && first == r) || (r.Length == 2 && second == r))            {                result += int.Parse(Enum.Format(R, Enum.Parse(R, r), d));                int lenR = r.Length, lenS = s.Length;                if (lenS - lenR < 1)                    s = ;                else                    s = s.Substring(lenR, lenS - lenR);                if (s.Length > 1)                {                    first = s.Substring(0, 1); second = s.Substring(0, 2);                }                else if (s.Length == 1)                {                    first = s.Substring(0, 1); second = ;                }                else                {                    first = ; second = ;                }            }        }        return result;    }}public enum Roman{    M = 1000,    CM = 900,    D = 500,    CD = 400,    C = 100,    XC = 90,    L = 50,    XL = 40,    X = 10,    IX = 9,    V = 5,    IV = 4,    I = 1};

You can run it ...... However, the efficiency is too low, so you can't bear to look directly ...... So I learned from other gods ......

The following code deeply touched me ...... Array as the index of the array ...... My least common usage ......

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

This algorithm makes full use of the relationship between the order of the two numbers, that is, if I is in front of V, that is, IV, it represents 4, and vice versa, it represents 6. Combined with C ++'s unordered_map, you can use arrays to judge and add or subtract ret.