13. 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，详细解释了罗马数字的构成原则及特殊情况处理，如IV表示4，IX表示9等。通过具体实例展示了如何从左到右解析罗马数字，以及如何在特殊情况下进行减法操作。

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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.

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Code:

class Solution {
    private static int map(char c) {
        switch (c) {
            case 'I': return 1;
            case 'V': return 5;
            case 'X': return 10;
            case 'L': return 50;
            case 'C': return 100;
            case 'D': return 500;
            default: return 1000;
        }
    }
    public int romanToInt(String s) {
        int result = 0,sign = 0,i = 0;
        for (; i < s.length() - 1; i++)
            if (s.charAt(i) == s.charAt(i + 1)) {
                result += map(s.charAt(i));
            } else if ((sign = -map(s.charAt(i)) + map(s.charAt(i + 1))) > 0) {
                result += sign;
                i++;
                sign = 0;
            } else
                result += map(s.charAt(i));
        return result + (s.length() == i ? 0 : map(s.charAt(s.length() - 1)));
    }
}