13. Roman to Integer

Description: 

https://leetcode.com/problems/roman-to-integer/#/description

Boundary:

s.size() = 0

Algorithm:

Use unordered_map m<char, int> to construct the correspondence between character and number like:

character	numerical value
I	1
V	5
X	10
L	50
C	100
D	500
M	1000

Then check the string s one by one

if m[s[i]] < m[s[i+1]]   result -= m[s[i]]

else   result += m[s[i]]

Code:

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

};

Tips:

Beats 85% , however
If I initialize m before considering boundary. It will be much slower (only beats 15%)

class Solution {
public:
int romanToInt(string s) {

m[‘I’] = 1;
m[‘V’] = 5;
m[‘X’] = 10;
m[‘L’] = 50;
m[‘C’] = 100;
m[‘D’] = 500;
m[‘M’] = 1000;
if (s.size() == 0) return 0;

int result = 0;
for (int i = 0; i < s.size() – 1; i++)
{
if (m[s[i]] < m[s[i + 1]])
result -= m[s[i]];
else
result += m[s[i]];
}
result += m[s[s.size() – 1]];
return result;
}
private:
unordered_map <char, int> m;

};