LeetCode12——Integer to Roman

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题目链接

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，

代码

class Solution:
    def intToRoman(self, num: 'int') -> 'str':
        res = ""
        carry = 1
        while num > 0:
            temp = num % 10
            num = num // 10
            if temp == 4 and carry == 1:
                res = "IV" + res
            elif temp == 9 and carry == 1:
                res = "IX" + res
            elif temp == 4 and carry == 10:
                res = "XL" + res
            elif temp == 9 and carry == 10:
                res = "XC" + res
            elif temp == 4 and carry == 100:
                res = "CD" + res
            elif temp == 9 and carry == 100:
                res = "CM" + res
            elif carry == 1 and temp < 5:
                res = ("".rjust(temp, "I")) + res
            elif carry == 1 and temp >= 5:
                res = ("V" + "".rjust(temp-5, "I")) + res
            elif carry == 10 and temp < 5:
                res = ("".rjust(temp,"X")) + res
            elif carry == 10 and temp >= 5:
                res = ("L" + "".rjust((temp-5),"X")) + res
            elif carry == 100 and temp < 5:
                res = ("".rjust(temp, "C")) + res
            elif carry == 100 and temp >= 5:
                res = ("D" + "".rjust((temp-5),"C")) + res
            elif carry == 1000:
                res = ("".rjust(temp, "M")) + res

            carry = carry * 10

        return res

按照题目意思，一步一步来就好，分情况处理，对每一位数分别进行处理，不过这么写，代码的可扩展性很差，太多的判断语句了。