LeetCode13-Roman to Integer(C++)

Description

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol Value
I                 1
V                5
X                10
L                50
C                100
D                500
M                1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:
Input: "III"
Output: 3

Example 2:
Input: "IV"
Output: 4

Example 3:
Input: "IX"
Output: 9

Example 4:
Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:
Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

AC代码

class Solution {
private:
    vector<pair<const char*, int> > conv = {
        {"M", 1000},
        {"D", 500},
        {"C", 100},
        {"L", 50},
        {"X", 10},
        {"V", 5},
        {"I", 1}
    };
public:
    int romanToInt(string s) {
        int res = 0; // 存储返回值
        int pos = 0; // 存储当前值
        int temp = 0; // 存储前一个值
        
        for(int i=0; i<s.size(); i++) {
            char cur = s[i];
            switch (cur) {
                case 'M': pos = 1000;
                    break;
                case 'D': pos = 500;
                    break;
                case 'C': pos = 100;
                    break;
                case 'L': pos = 50;
                    break;
                case 'X': pos = 10;
                    break;
                case 'V': pos = 5;
                    break;
                case 'I': pos = 1;
                    break;
                default:
                    break;
            }

            res += pos;
            if(temp < pos) {  // 前一个值小于当前值，类似于CM, IV这种情况
                res -= 2*temp;
            }
            temp = pos;
        }
        return res;
    }
};

测试代码

int main() {
    Solution s;
    string a1 = "III";
    string a2 = "IV";
    string a3 = "IX";
    string a4 = "LVIII";
    string a5 = "MCMXCIV";
    cout << a1 << " string to int -> " << s.romanToInt(a1) << endl;
    cout << a2 << " string to int -> " << s.romanToInt(a2) << endl;
    cout << a3 << " string to int -> " << s.romanToInt(a3) << endl;
    cout << a4 << " string to int -> " << s.romanToInt(a4) << endl;
    cout << a5 << " string to int -> " << s.romanToInt(a5) << endl;
}

总结

本题的关键是如何处理"CM, IX..."这类情况, 这类情况有一个共同的特点, 就是下一个罗马字符大于上一个, 所以我们可以设置两个flag变量, 一个用来记录当前的字符，另外一个用户来记录前一个字符，如果出现CM这种情况的话, 就+M-2*C。