letecode [13] - Roman to Integer

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.

题目大意：根据输入的罗马数字，输出规则对应的相应整数。

理  解 ：遍历输入的罗马数字，转化为整数值存于数组中。

　　　　遍历数组，若当前值比右边数大，则累加为和，i = i + 1；反之，则用右边的数-当前值，i = i + 2。（注意边界值的处理）

　　　　别人的解法：遍历罗马数字时，使用if进行比较，直接计算。这种效率从当前题目上看会高一点点，可不计。

代  码 C++：

class Solution {
public:
    int romanToInt(string s) {
        int n = s.length();
        int *arr = new int[n];
        int i,value=0;
        
        for(i=0;i<n;++i){
            switch(s[i]){
                case ‘I‘:
                    arr[i] = 1;
                    break;
                case ‘V‘:
                    arr[i] = 5;
                    break;
                case ‘X‘:
                    arr[i] = 10;
                    break;
                case ‘L‘:
                    arr[i] = 50;
                    break;
                case ‘C‘:
                    arr[i] = 100;
                    break;
                case ‘D‘:
                    arr[i] = 500;
                    break;
                case ‘M‘:
                    arr[i] = 1000;
                    break;
            }
        }
        
        for(i=0;i<n;++i){
            if(i==n-1 || arr[i]>=arr[i+1]){
                value = value + arr[i];
            }else{
                value = value + arr[i+1] - arr[i];
                i++;
            }
        }
        return value;
    }
};