13. Roman to [number to integer Rome] Integer

Description
character value
the I. 1
V. 5
X-10
L 50
C 100
D 500
M 1000

I may be on the left V (5) and X (10), and to represent 4 and 9.
X L can be placed on the left (50) and C (100), and 40 and 90 are represented.
C may be placed on D (500) and M (1000) to the left, to 400 and 900 represent.
Given a Roman numeral, to convert it to an integer. To ensure that the input is in the range of 1 to 3999

example

Thinking

• switch faster than some more dict
• IV IX XL XC CD CM centralized exceptional circumstances, if the current is less than the Roman numerals behind the Roman numerals, on subtracting the current Roman numerals, Roman numerals or plus the current
  answer
• python (no switch)

class Solution:
    def romanToInt(self, s: str) -> int:
        d={'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
        num=0
        for i in range(len(s)):
            if i+1<=len(s)-1 and d[s[i+1]]>d[s[i]]:
                num -= d[s[i]] 
            else:
                num += d[s[i]]
        return num

• c++

* 用字典

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

        int num=0;
        int now=0;
        int pre=0;
        for (int i=0; i<s.size(); i++)
        {
            switch(s[i]) {
                case 'I': now = 1; break;
                case 'V': now = 5; break;
                case 'X': now = 10; break;
                case 'L': now = 50; break;
                case 'C': now = 100; break;
                case 'D': now = 500; break;
                case 'M': now = 1000; break;
                
            }
            
            if (pre<now) num-=pre*2;
            num+=now;
            pre=now;
            
        }
        
        return num;
    }
};