LeetCode Roman to Integer

原题链接在这里：https://leetcode.com/problems/roman-to-integer/

取当前数值若是比之前的小，就正常加上当前数值，若是比之前位大说明刚才加错了，需要先减掉在加上(cur-pre).

e.g. "XIV", at the beginning, res = 10. When i = 1, cur = 1, pre = 10, cur<pre, res = 11; when i = 2, cur = 5, pre = 1, cur>pre, res = 11-1+(5-1) = 14. 若是cur==pre，正常加cur，"XX" = 20.

AC Java:

public class Solution {
    public int romanToInt(String s) {
        int res = charToInt(s.charAt(0));
        for(int i = 1; i < s.length(); i++){
            int cur = charToInt(s.charAt(i));
            int pre = charToInt(s.charAt(i-1));
            if(cur>pre){
                res = res-pre+(cur-pre);
            }else{
                res+=cur;
            }
        }
        return res;
    }
    
    private int charToInt(char c){
        int res = 0;
        switch(c){
            case 'I':
                res = 1;
                break;
            case 'V':
                res = 5;
                break;
            case 'X':
                res = 10;
                break;
            case 'L':
                res = 50;
                break;
            case 'C':
                res = 100;
                break;
            case 'D':
                res = 500;
                break;
            case 'M':
                res = 1000;
                break;
            default:
                res = 0;
        }
        return res;
    }
}