力扣题:数字与字符串间转换-12.21

力扣题-12.21

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：13. 罗马数字转整数

解题思想：进行遍历计算整数即可

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

class Solution {
public:
    int romanToInt(string s) {
        std::unordered_map<char, int> dic = {{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
        int num = dic[s[0]];
        int temp = dic[s[0]];

        for (int i = 1; i < s.length(); ++i) {
            int current = dic[s[i]];

            if (current <= temp) {
                num += current;
                temp = current;
            } else {
                num = num + current - 2 * temp;
                temp = current;
            }
        }

        return num;
    }
};