Leetcode-013-罗马数字转整数

Posted 2020-11-26 huangzengrui

双指针问题，当前的数小于后一位，就减去它的值，否则就加上它的值。

class Solution {
    public int romanToInt(String s) {
        Map<Character, Integer> demo = new HashMap<>();
        demo.put(‘I‘, 1);
        demo.put(‘V‘, 5);
        demo.put(‘X‘, 10);
        demo.put(‘L‘, 50);
        demo.put(‘C‘, 100);
        demo.put(‘D‘, 500);
        demo.put(‘M‘, 1000);

        int result = 0;
        for(int i=0;i<s.length()-1;i++){
            if(demo.get(s.charAt(i))<demo.get(s.charAt(i+1))){
                result -= demo.get(s.charAt(i));
            }else{
                result += demo.get(s.charAt(i));
            }
        }
        result += demo.get(s.charAt(s.length()-1));
        return result;
    }
}

class Solution:
    def romanToInt(self, s: str) -> int:
        demo = {‘I‘:1, ‘V‘:5, ‘X‘:10, ‘L‘:50, ‘C‘:100, ‘D‘:500, ‘M‘:1000}

        result = 0

        for i in range(len(s)-1):
            if demo[s[i]] < demo[s[i+1]]:
                result -= demo[s[i]]
            else:
                result += demo[s[i]]
        result += demo[s[len(s)-1]]
        return result