LeetCode重点题系列之【12. Integer to Roman】

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本文介绍了一种将整数转换为罗马数字的方法，并提供了一个具体的Python实现案例。文章详细解释了罗马数字的基本构成及其特殊组合规则，例如4和9的表示方式。

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: L = 50, V = 5, III = 3.

Example 5:

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

The trick is that add 4, 900 and other candidates into the dictionary. Use divmod function to calculate how many times we should use according symbol, and update the number.

class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        dic={1000:"M",900:"CM",500:"D",400:"CD",100:"C",90:"XC",50:"L",40:"XL",10:"X",9:"IX",5:"V",4:"IV",1:"I"}
        res=""
        for key in sorted(dic,reverse=True):
            q,r=divmod(num,key)
            if q>0:
                res+=q*dic[key]
                num=r
            if num==0:
                return res