leetcode: Roman to Integer

http://oj.leetcode.com/problems/roman-to-integer/

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

思路：

根据个十百千位分别作为一个状态机处理就可以了。

class Solution {
public:
    int romanToInt(string s) {
        int val = 0, mul = 0;
        char one, five, ten;
        const char *p = s.c_str();
        
        while (*p != '') {
            char ch = *p;
            
            if ('M' == ch) {
                mul = 1000;
                one = 'M';
                five = '?';
                ten = '?';
            }
            else if (('C' == ch) || ('D' == ch)) {
                mul = 100;
                one = 'C';
                five = 'D';
                ten = 'M';
            }
            else if (('X' == ch) || ('L' == ch)) {
                mul = 10;
                one = 'X';
                five = 'L';
                ten = 'C';
            }
            else {  // 'I' or 'V'.
                mul = 1;
                one = 'I';
                five = 'V';
                ten = 'X';
            }
            
            int tmp;
            
            if (one == ch) {
                tmp = 1;
                ++p;
                while ((*p != '') && (one == *p)) {
                    ++tmp;
                    ++p;
                }
                
                if (ten == *p) {
                    val += (mul * 9);
                    ++p;
                }
                else if (five == *p) {
                    val += (mul * 4);
                    ++p;
                }
                else {
                    val += (mul * tmp);
                }
            }
            else if (five == ch) {
                tmp = 5;
                ++p;
                while ((*p != '') && (one == *p)) {
                    ++tmp;
                    ++p;
                }
                
                val += (mul * tmp);
            }
            // No need to handle ten, because it will be one in next level.
        }
        
        return val;
    }
};