integer-to-roman.py

"""

12. Integer to Roman
Medium


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, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 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.

 

Example 1:

Input: num = 3
Output: "III"
Explanation: 3 is represented as 3 ones.
Example 2:

Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 3:

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

Constraints:

1 <= num <= 3999

"""

# V0 
# IDEA
# In [8]: 1200%100
# Out[8]: 0
# In [9]: 1234%1000
# Out[9]: 234
# In [10]: 1234%100
# Out[10]: 34
# In [11]: 1234%10
# Out[11]: 4
# example :
# def str_2_int(x):
#     r=0
#     for i in x:
#         r = int(r)*10 + int(i)
#         print (i, r)
#     return r
class Solution(object):
    def intToRoman(self, num):
        M = ["", "M", "MM", "MMM"]
        C = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"]
        X = ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"]
        I = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"]
        return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10]

# V1
# https://blog.csdn.net/NXHYD/article/details/72461202
class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        c = [['','I','II','III','IV','V','VI','VII','VIII','IX'],
             ['','X','XX','XXX','XL','L','LX','LXX','LXXX','XC'],
             ['','C','CC','CCC','CD','D','DC','DCC','DCCC','CM'],
             ['','M','MM','MMM']]
        s = []
        i = 3
        while num :
            s.append(c[i][num/pow(10, i)])
            num %= pow(10, i)
            i -= 1
        return ''.join(s)

# V1
# IDEA : Greedy
# https://leetcode.com/problems/integer-to-roman/solution/
class Solution:
    def intToRoman(self, num: int) -> str:
        digits = [(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")]
        
        roman_digits = []
        # Loop through each symbol.
        for value, symbol in digits:
            # We don't want to continue looping if we're done.
            if num == 0: break
            count, num = divmod(num, value)
            # Append "count" copies of "symbol" to roman_digits.
            roman_digits.append(symbol * count)
        return "".join(roman_digits)

# V1
# IDEA : Hardcode Digits
# https://leetcode.com/problems/integer-to-roman/solution/
class Solution:
    def intToRoman(self, num: int) -> str:
        thousands = ["", "M", "MM", "MMM"]
        hundreds = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"]
        tens = ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"]
        ones = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"]
        return (thousands[num // 1000] + hundreds[num % 1000 // 100] 
               + tens[num % 100 // 10] + ones[num % 10])

# V1' 
# https://blog.csdn.net/NXHYD/article/details/72461202
class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        M = ["", "M", "MM", "MMM"]
        C = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"]
        X = ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"]
        I = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"]
        return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10]

# V1'' 
# https://www.jiuzhang.com/solution/integer-to-roman/#tag-highlight-lang-python
class Solution:
    def parse(self, digit, index):
        NUMS = {
            1: 'I',
            2: 'II',
            3: 'III',
            4: 'IV',
            5: 'V',
            6: 'VI',
            7: 'VII',
            8: 'VIII',
            9: 'IX',
        }
        ROMAN = {
            'I': ['I', 'X', 'C', 'M'],
            'V': ['V', 'L', 'D', '?'],
            'X': ['X', 'C', 'M', '?']
        }
        
        s = NUMS[digit]
        return s.replace('X', ROMAN['X'][index]).replace('I', ROMAN['I'][index]).replace('V', ROMAN['V'][index])

    # @param {integer} num
    # @return {string}        
    def intToRoman(self, num):
        s = ''
        index = 0
        while num != 0:
            digit = num % 10
            if digit != 0:
                s = self.parse(digit, index) + s
            num = num / 10
            index += 1
        return s

# V2 
# https://blog.csdn.net/NXHYD/article/details/72461202
class Solution(object):
    def intToRoman(self, num):
        import math 
        """
        :type num: int
        :rtype: str
        """
        M = ["", "M", "MM", "MMM"]
        C = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"]
        X = ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"]
        I = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"]
        return M[(math.floor(num/1000))] + C[(math.floor(num%1000)/100)] + X[(math.floor(num%100)/10)] + I[(math.floor(num%10))]


# V3
# Time:  O(n)
# Space: O(1)
class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        numeral_map = {1: "I", 4: "IV", 5: "V", 9: "IX", \
                       10: "X", 40: "XL", 50: "L", 90: "XC", \
                       100: "C", 400: "CD", 500: "D", 900: "CM", \
                       1000: "M"}
        keyset, result = sorted(numeral_map.keys()), []

        while num > 0:
            for key in reversed(keyset):
                while num / key > 0:
                    num -= key
                    result += numeral_map[key]

        return "".join(result)