LeetCode 每日一题 13.罗马数字转整数

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问题描述

简要思路

简单题，暴力 if。
简洁点的用map。

代码

class Solution {
    
    
public:
    int romanToInt(string s) {
    
    
        int n = s.length();
        int num = 0;
        for(int i=0; i<n; i++){
    
    
            if(s[i] == 'M') num += 1000;
            else if(s[i] == 'C' && s[i+1] == 'M'){
    
    
                num += 900;
                i++;
            }
            else if(s[i] == 'C' && s[i+1] == 'D'){
    
    
                num += 400;
                i++;
            }
            else if(s[i] == 'X' && s[i+1] == 'C'){
    
    
                num += 90;
                i++;
            }
            else if(s[i] == 'X' && s[i+1] == 'L'){
    
    
                num += 40;
                i++;
            }
            else if(s[i] == 'I' && s[i+1] == 'X'){
    
    
                num += 9;
                i++;
            }
            else if(s[i] == 'I' && s[i+1] == 'V'){
    
    
                num += 4;
                i++;
            }
            else if(s[i] == 'D') num += 500;
            else if(s[i] == 'C') num += 100;
            else if(s[i] == 'L') num += 50;
            else if(s[i] == 'X') num += 10;
            else if(s[i] == 'V') num += 5;
            else if(s[i] == 'I') num += 1;
        }

        return num;
    }
};