leetcode 12. Integer to Roman【Python】模拟

版权声明：⁄(⁄ ⁄•⁄ω⁄•⁄ ⁄)⁄hiahiahia 欢迎斧正 https://blog.csdn.net/zhou_yujia/article/details/81279494

---

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.

字符串转化成罗马字符

开始写了一个不是for循环的 写了一半就20来行了 把自己恶心到了 就此作罢

多写了两个字符是为了防止报错，可以统一处理

class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        ans=""
        ch1=['I','X','C','M','H']
        ch5=['V','L','D','Q']
        for i in range(3,-1,-1):
            t=num%(10**(i+1))/(10**i)
            if(t==9):
                ans=ans+ch1[i]+ch1[i+1]
            elif(t==4):
                ans=ans+ch1[i]+ch5[i]
            elif(t>=5):
                t=t-5
                ans=ans+ch5[i]
            if(t<=3):
                for j in range(t):
                    ans=ans+ch1[i]
        return ans