Leetcode013-Roman to Integer

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本文介绍了一种将罗马数字转换为整数的方法。通过使用数组来对应不同的罗马符号，并利用特殊情况的处理，实现了一个高效的转换算法。文章提供了一个具体的C++代码示例。

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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.

最开始想用每种情况都写一遍，虽然会的不多，但还是嫌麻烦，想到了Python中的用两个list去对应相同的索引的不同类型，所以用数组老对应不同的罗马符号。

Solution：

class Solution {
public:
    int romanToInt(string s) {
        int num[] = {1, 5, 10, 50, 100, 500, 1000};
        string rom = "IVXLCDM";
        vector<int> tmp;
        int res = 0;
        for (int i = 0; i < s.size(); i++)
        {
            for (int j = 0; j < 7; j++)
            {
                if (s[i] == rom[j])
                    tmp.push_back(num[j]);
            }
        }
        for (int i = 0; i < tmp.size(); i++)
        {
            if (i == tmp.size() - 1 || tmp[i] >= tmp[i + 1])
            {
                res += tmp[i];
            }
            else
            {
                res -= tmp[i];
            }
        }
        return res;
    }
};