leetcode 12 -> Integer to Roman

转载于 2019-03-10 19:40:00 发布 · 75 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/sea-stream/p/10506657.html

本文介绍了一个将整数转换为罗马数字的递归算法实现。通过一系列条件判断，该算法能够根据输入的整数值返回对应的罗马数字字符串。文章详细展示了算法的逻辑流程，包括对千位、百位、十位和个位数的处理。

class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        if num == 0:
            return ""
        if num >= 1000:
            return "M" + self.intToRoman(num - 1000)
        elif num >= 900:
            return "CM" + self.intToRoman(num - 900)
        elif num >= 500:
            return "D" + self.intToRoman(num - 500)
        elif num >= 400:
            return "CD" + self.intToRoman(num - 400)
        elif num >= 100:
            return "C" + self.intToRoman(num - 100)
        elif num >= 90:
            return "XC" + self.intToRoman(num - 90)
        elif num >= 50:
            return "L" + self.intToRoman(num - 50)
        elif num >= 40:
            return "XL" + self.intToRoman(num - 40)
        elif num >= 10:
            return "X" + self.intToRoman(num - 10)
        elif num >= 9:
            return "IX" + self.intToRoman(num - 9)
        elif num >= 5:
            return "V" + self.intToRoman(num - 5)
        elif num >= 4:
            return "IV" + self.intToRoman(num - 4)
        elif num >= 1:
            return "I" + self.intToRoman(num - 1)