leetcode-algorithms-13 Roman to Integer

leetcode-algorithms-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.
• X can be placed before L (50) and C (100) to make 40 and 90.

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.

解法

观察罗马数字,其对应的值如果前一位数比后一个位数小就是减掉去值,否由增加值.

例: IV = ?,I = 1,V = 5,则,IV = -I + V. VI = +V + I.

class Solution
{
public:
    int romanToInt(string s)
    {
        int num[20];
        int size = s.size();
        int sum = 0;
        for (int i = 0; i < size; ++i)
        {
            switch(s[i]) {
                case 'M':
                    num[i] = 1000;
                    break;
                case 'D':
                    num[i] = 500;
                    break;
                case 'C':
                    num[i] = 100;
                    break;
                case 'L':
                    num[i] = 50;
                    break;
                case 'X':
                    num[i] = 10;
                    break;
                case 'V':
                    num[i] = 5;
                    break;
                case 'I':
                    num[i] = 1;
                    break;
            }
            if (i > 0 && num[i - 1] < num[i])
                sum -= num[i - 1];
            else if (i > 0)
                sum += num[i - 1];
            if (i == size - 1) sum += num[i];
        }

        return sum;
    }
};

时间复杂度: O(n).

空间复杂度: O(20);

链接: leetcode-algorithms 目录