1、题目

12. Integer to Roman

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.

2、我的解答（未简化）

 # -*- coding: utf-8 -*-

 # @Time    : 2020/1/30 19:14

 # @Author  : SmartCat0929

 # @Email   : 1027699719@qq.com

 # @Link    : https://github.com/SmartCat0929

 # @Site    :

 # @File    : 12. Integer to Roman(自己想的复杂的算法）.py

 class Solution:

     def intToRoman(self, num: int) -> str:

         numStr = str(num)

         if num < 10:

             if int(numStr[-1]) > 0 and int(numStr[-1]) < 4:

                 units = ""

                 for i in range(0, int(numStr[-1])):

                     units = "I" + units

             elif int(numStr[-1]) == 4:

                 units = "IV"

             elif int(numStr[-1]) == 5:

                 units = "V"

             elif int(numStr[-1]) > 5 and int(numStr[-1]) < 9:

                 units = "V"

                 for i in range(5, int(numStr[-1])):

                     units = units + "I"

             elif int(numStr[-1]) == 9:

                 units = "IX"

             return units

         elif num >= 10 and num <= 99:  # 如果是两位数

             if int(numStr[-1]) == 0:

                 units = ""

             elif int(numStr[-1]) > 0 and int(numStr[-1]) < 4:  # 个位

                 units = ""

                 for i in range(0, int(numStr[-1])):

                     units = "I" + units

             elif int(numStr[-1]) == 4:

                 units = "IV"

             elif int(numStr[-1]) == 5:

                 units = "V"

             elif int(numStr[-1]) > 5 and int(numStr[-1]) < 9:

                 units = "V"

                 for i in range(5, int(numStr[-1])):

                     units = units + "I"

             elif int(numStr[-1]) == 9:

                 units = "IX"

             if int(numStr[-2]) == 1:  # 十位

                 decades = "X"

             elif int(numStr[-2]) > 0 and int(numStr[-2]) < 4:  # 十位

                 decades = ""

                 for j in range(0, int(numStr[-2])):

                     decades = "X" + decades

             elif int(numStr[-2]) == 4:

                 decades = "XL"

             elif int(numStr[-2]) == 5:

                 decades = "L"

             elif int(numStr[-2]) > 5 and int(numStr[-2]) < 9:

                 decades = "L"

                 for j in range(5, int(numStr[-2])):

                     decades = decades + "X"

             elif int(numStr[-2]) == 9:

                 decades = "XC"

             romans = decades + units

             return romans

         elif num >= 100 and num <= 999:  # 如果是三位数

             if int(numStr[-1]) == 0:  # 个位

                 units = ""

             elif int(numStr[-1]) > 0 and int(numStr[-1]) < 4:

                 units = ""

                 for i in range(0, int(numStr[-1])):

                     units = "I" + units

             elif int(numStr[-1]) == 4:

                 units = "IV"

             elif int(numStr[-1]) == 5:

                 units = "V"

             elif int(numStr[-1]) > 5 and int(numStr[-1]) < 9:

                 units = "V"

                 for i in range(5, int(numStr[-1])):

                     units = units + "I"

             elif int(numStr[-1]) == 9:

                 units = "IX"

             if int(numStr[-2]) == 0:  # 十位

                 decades = ""

             elif int(numStr[-2]) > 0 and int(numStr[-2]) < 4:

                 decades = ""

                 for j in range(0, int(numStr[-2])):

                     decades = "X" + decades

             elif int(numStr[-2]) == 4:

                 decades = "XL"

             elif int(numStr[-2]) == 5:

                 decades = "L"

             elif int(numStr[-2]) > 5 and int(numStr[-2]) < 9:

                 decades = "L"

                 for j in range(5, int(numStr[-2])):

                     decades = decades + "X"

             elif int(numStr[-2]) == 9:

                 decades = "XC"

             if int(numStr[-3]) > 0 and int(numStr[-3]) < 4:  # 百位

                 hundreds = ""

                 for j in range(0, int(numStr[-3])):

                     hundreds = "C" + hundreds

             elif int(numStr[-3]) == 4:

                 hundreds = "CD"

             elif int(numStr[-3]) == 5:

                 hundreds = "D"

             elif int(numStr[-3]) > 5 and int(numStr[-3]) < 9:

                 hundreds = "D"

                 for j in range(5, int(numStr[-3])):

                     hundreds = hundreds + "C"

             elif int(numStr[-3]) == 9:

                 hundreds = "CM"

             romans = hundreds + decades + units

             return romans

         elif num >= 1000 and num <= 3999:  # 如果是四位数

             if int(numStr[-1]) == 0:  # 个位

                 units = ""

             elif int(numStr[-1]) > 0 and int(numStr[-1]) < 4:

                 units = ""

                 for i in range(0, int(numStr[-1])):

                     units = "I" + units

             elif int(numStr[-1]) == 4:

                 units = "IV"

             elif int(numStr[-1]) == 5:

                 units = "V"

             elif int(numStr[-1]) > 5 and int(numStr[-1]) < 9:

                 units = "V"

                 for i in range(5, int(numStr[-1])):

                     units = units + "I"

             elif int(numStr[-1]) == 9:

                 units = "IX"

             if int(numStr[-2]) == 0:  # 十位

                 decades = ""

             elif int(numStr[-2]) > 0 and int(numStr[-2]) < 4:

                 decades = ""

                 for j in range(0, int(numStr[-2])):

                     decades = "X" + decades

             elif int(numStr[-2]) == 4:

                 decades = "XL"

             elif int(numStr[-2]) == 5:

                 decades = "L"

             elif int(numStr[-2]) > 5 and int(numStr[-2]) < 9:

                 decades = "L"

                 for j in range(5, int(numStr[-2])):

                     decades = decades + "X"

             elif int(numStr[-2]) == 9:

                 decades = "XC"

             if int(numStr[-3]) == 0:  # 百位

                 hundreds = ""

             elif int(numStr[-3]) > 0 and int(numStr[-3]) < 4:

                 hundreds = ""

                 for j in range(0, int(numStr[-3])):

                     hundreds = "C" + hundreds

             elif int(numStr[-3]) == 4:

                 hundreds = "CD"

             elif int(numStr[-3]) == 5:

                 hundreds = "D"

             elif int(numStr[-3]) > 5 and int(numStr[-3]) < 9:

                 hundreds = "D"

                 for j in range(5, int(numStr[-3])):

                     hundreds = hundreds + "C"

             elif int(numStr[-3]) == 9:

                 hundreds = "CM"

             if int(numStr[-4]) > 0 and int(numStr[-4]) < 4: #千位

                 thousands = ""

                 for j in range(0, int(numStr[-4])):

                     thousands = "M" + thousands

                 romans = thousands + hundreds + decades + units

                 return romans

         elif num > 3999:

             return 0

 print(Solution().intToRoman(400))

3、大神的解法

LeetCode上看到的大神的解法，实在是佩服！ 采用字典将特殊符号记住，随后逐一判断.......

 class Solution:

     def intToRoman(self, num: int) -> str:

         res = ""

         s = 1000

         d = {1: "I", 5: "V", 10: "X", 50: "L", 100: "C", 500: "D", 1000: "M"}

         while num != 0:

             r, temp = divmod(num, s)

             if r == 9:

                 res += d[s] + d[s * 10]

             elif r == 4:

                 res += d[s] + d[s * 5]

             elif r >= 5:

                 res += d[s * 5] + d[s] * (r - 5)

             else:

                 res += d[s] * r

             s = s // 10

             num = temp

         return res

 print(Solution().intToRoman(2964))
												


											
leetCode练题——12. Integer to Roman的更多相关文章
	

    
								
  1. Leetcode 12——Integer to Roman
		
    12.Integer to Roman Given an integer, convert it to a roman numeral. Input is guaranteed to be withi ...
    
		
    2. 
						
  2. Leetcode  12.   Integer to Roman(打表，水)
		
    12. Integer to Roman Medium Roman numerals are represented by seven different symbols: I, V, X, L, C ...
    
		
    3. 
						
  3. 乘风破浪：LeetCode真题_007_Reverse Integer
		
    乘风破浪:LeetCode真题_007_Reverse Integer 一.前言 这是一个比较简单的问题了,将整数翻转,主要考察了取整和取余,以及灵活地使用long型变量防止越界的问题. 二.Reve ...
    
		
    4. 
						
  4. 《LeetBook》leetcode题解(12):Integer to Roman[M]
		
    我现在在做一个叫<leetbook>的免费开源书项目,力求提供最易懂的中文思路,目前把解题思路都同步更新到gitbook上了,需要的同学可以去看看 书的地址:https://hk029.g ...
    
		
    5. 
						
  5. 【LeetCode】12. Integer to Roman (2 solutions)
		
    Integer to Roman Given an integer, convert it to a roman numeral. Input is guaranteed to be within t ...
    
		
    6. 
						
  6. 【算法】LeetCode算法题-Reverse Integer
		
    这是悦乐书的第143次更新,第145篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第2题(顺位题号是7),给定32位有符号整数,然后将其反转输出.例如: 输入: 123  ...
    
		
    7. 
						
  7. leetCode练题——13. Roman to Integer
		
    1.题目13. Roman to Integer Roman numerals are represented by seven different symbols: I, V, X, L, C, D ...
    
		
    8. 
						
  8. [LeetCode] 12. Integer to Roman 整数转化成罗马数字
		
    Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1  ...
    
		
    9. 
						
  9. 【LeetCode】12. Integer to Roman 整数转罗马数字
		
    作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:roman, 罗马数字,题解,leetcode, 力扣, ...
    
		
    10. 
		

	


随机推荐
	

    
									
  1. HDU - 5187 zhx's contest（快速幂+快速乘法）
			
    作为史上最强的刷子之一,zhx的老师让他给学弟(mei)们出n道题.zhx认为第i道题的难度就是i.他想要让这些题目排列起来很漂亮. zhx认为一个漂亮的序列{ai}下列两个条件均需满足. 1:a1. ...
    
			
    2. 
						
  2. opencv：形态学梯度
			
    #include <opencv2/opencv.hpp> #include <iostream> using namespace cv; using namespace st ...
    
			
    3. 
						
  3. vue+vuex项目中怎么实现input模糊查询
			
    1,首先给input框添加方法,但是用的是element-ui的组件,对input进行了封装,不能直接用原生的方法!,在element组件中,input框中方法有实例参数$event,代表事件对象   ...
    
			
    4. 
						
  4. python 网页中文显示Unicode码
			
    print repr(a).decode("unicode–escape") 注:a是要输出的结果,
    
			
    5. 
						
  5. Linux查看mysql是否启动的命令
			
    使用命令 # service mysqld status 或者 # service mysql status 来查看mysql 的启动状态. 如果是 mysqld is stopped 那就说明mys ...
    
			
    6. 
						
  6. Java生鲜电商平台-生鲜电商高并发下的接口幂等性实现与代码讲解
			
    Java生鲜电商平台-生鲜电商高并发下的接口幂等性实现与代码讲解 说明:Java生鲜电商平台-生鲜电商高并发下的接口幂等性实现与代码讲解,实际系统中有很多操作,是不管做多少次,都应该产生一样的效果或返 ...
    
			
    7. 
						
  7. JAVA 中集合ConcurrentMap
			
    ConcurrentMap ConcurrentMap,它是一个接口,是一个能够支持并发访问的java.util.map集合 在原有java.util.map接口基础上又新提供了4种方法,进一步扩展了 ...
    
			
    8. 
						
  8. stream.js
			
    <script src='stream-min.js'></script> 下载 stream.js 2Kb minified streams是什么? Streams 是一个操 ...
    
			
    9. 
						
  9. 一张linux光盘查看是哪个版本号的方法
			
    在此查看版本号,方法如下:打开光盘,查找rpm包中的release,就是版本号.
    
			
    10. 
						
  10. HDU4081 Qin Shi Huang's National Road System
			
    先求最小生成树 再遍历每一对顶点,如果该顶点之间的边属于最小生成树,则剪掉这对顶点在最小生成树里的最长路径 否则直接剪掉连接这对顶点的边~ 用prim算法求最小生成树最长路径的模板~ #include ...
    
			
    11.