Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

I:1

V:5

X:10

L:50

C:100

D:500

M:1000

比如：MCMLXX：1970

MDCLXVI:1666

思路：從高位到低位掃描

可將1970劃分為

[M] [CM] [L] [X] [X]，即每個段內都是非降序，程式上體現為當出現升序時，減去前一個數字，否則加上前一個數字

class Solution {
    public int measure(char c){
        if(c=='I'){
            return 1;
        }else if(c=='V'){
            return 5;
        }else if(c=='X'){
            return 10;
        }else if(c=='L'){
            return 50;
        }else if(c=='C'){
            return 100;
        }else if(c=='D'){
            return 500;
        }else if(c=='M'){
            return 1000;
        }
        return 0;
    }
    
    public int romanToInt(String s) {
        int sum = 0;
        int pre = 0;
        int temp = 0;
        for(int i = 0;i <s.length() ;i++){
            temp = measure(s.charAt(i));
            if(temp <= pre){
                sum += pre;
                pre = temp;
            }else{
                sum -= pre;
                pre = temp;
            }
        }
        sum += pre;
        return sum;
    }
}