leetcode12. Integer to Roman

问题描述：

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 an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: "III"

Example 2:

Input: 4
Output: "IV"

Example 3:

Input: 9
Output: "IX"

Example 4:

Input: 58
Output: "LVIII"
Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

 

参考：

http://www.cnblogs.com/grandyang/p/4123374.html

作者一共提供了3种方法，最后一种达到了O(1)

 

思路：

说到罗马数字，每个字母代表不同的数值，且都为整数值，所以用个字典存储起来可以方便提取，而且

在这个方法中，对每个位置可能出现的情况进行了详细的划分：

0-3: 为了保证为0时不出现字母，则选择从循环从1开始

4: 为特殊形式所以特殊处理

5-8：大加小

9:大减小

对每种情况都进行讨论。

 

代码：

class Solution {
public:
    string intToRoman(int num) {
        string ret = "";
        char roman[] = {'M', 'D', 'C', 'L', 'X', 'V', 'I'};
        int nums[] = {1000, 500, 100, 50, 10, 5, 1};
        for(int i = 0; i < 7; i += 2){
            int v = num / nums[i];
            if(v < 4){
                for(int j = 1; j <= v; j++)
                    ret += roman[i];
            }else if(v == 4){
                ret = ret + roman[i] + roman[i-1];
            }else if(v > 4 && v < 9){
                ret += roman[i-1];
                for(int j = 5; j < v; j++)
                    ret += roman[i];
            }else if(v == 9){
                ret = ret + roman[i] + roman[i-2];
            }
            num %= nums[i];
        }
        return ret;
    }
};