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

For example, two is written as IIin 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.

Example:

Input: "III" Output: 3
Input: "IV" Output: 4
Input: "IX" Output: 9
Input: "LVIII" Output: 58 Explanation: C = 100, L = 50, XXX = 30 and III = 3.
Input: "MCMXCIV" Output: 1994 Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

翻译：

将用罗马字表达式翻译成阿拉伯数字即可。

1. 把特殊的值挑出来即可

实际耗时：48ms

 public int romanToInt(String s) {
        int length = s.length();
        int sum = 0;
        for (int i = 0; i < length; i++) {
            char a = s.charAt(i);
            if (i < length - 1) {
                char b = s.charAt(i + 1);
                if (a == 'I' && b == 'V') {
                    sum += 4;
                    i++;
                    continue;
                } else if (a == 'I' && b == 'X') {
                    sum += 9;
                    i++;
                    continue;
                } else if (a == 'X' && b == 'L') {
                    sum += 40;
                    i++;
                    continue;
                } else if (a == 'X' && b == 'C') {
                    sum += 90;
                    i++;
                    continue;
                } else if (a == 'C' && b == 'D') {
                    sum += 400;
                    i++;
                    continue;
                } else if (a == 'C' && b == 'M') {
                    sum += 900;
                    i++;
                    continue;
                }
            }

            if (a == 'I') {
                sum += 1;
            }
            if (a == 'V') {
                sum += 5;
            }
            if (a == 'X') {
                sum += 10;
            }
            if (a == 'L') {
                sum += 50;
            }
            if (a == 'C') {
                sum += 100;
            }
            if (a == 'D') {
                sum += 500;
            }
            if (a == 'M') {
                sum += 1000;
            }
        }
        return sum;
    }

就是简单把那些4 9 40 90这种数字挑出来，其余就加起来即可。

时间复杂度O(n) (n为数字长度)

空间复杂度O(1)

2. 其他方法

实际耗时：未知

暂时没有写

其实这种规定好了的规则下，还是用手工把所有情况都“遍历”一遍来的最快，其他无非就是把上面的规则翻译一下放进数组或者hashmap中，并不会加快算法效率。

时间复杂度O(未知)

空间复杂度O(未知)