LeetCode 012 Integer to Roman

LeetCode

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

题目大意：

阿拉伯数字转罗马数字

解题思路(推荐)：

本质上和算法二一样，但优化了代码。map的内容是一样的但变成list保证顺序，然后从大到小遍历这个map，商对应的symbol放入结果，余数进入下一轮

注意事项：

1. map的内容是一样的但变成list保证顺序，然后从大到小遍历这个map，商对应的symbol放入结果，余数进入下一轮

Python代码：

def intToRoman(self, num: int) -> str:
	INT_TO_ROMAN = [(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')]
	res = ''
	for n, symbol in INT_TO_ROMAN:
		count, num = num // n, num % n
		res += symbol * count
	return res

算法分析：

时间复杂度为O(1)，空间复杂度O(1)

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算法II解题思路：

N/A

解题步骤：

N/A

注意事项：

1. 用int to english的递归方法，将固定值放入到Map中
2. 分界点为[4, 5, 9 10], [40, 50, 90, 100]进行递归

Python代码：

def intToRoman2(self, num: int) -> str:
	INT_TO_ROMAN = {
		0: '',
		4: 'IV',
		5: 'V',
		9: 'IX',
		10: 'X',
		40: 'XL',
		50: 'L',
		90: 'XC',
		100: 'C',
		400: 'CD',
		500: 'D',
		900: 'CM',
		1000: 'M',
	}
	if num in INT_TO_ROMAN:
		return INT_TO_ROMAN[num]
	elif num < 4:
		return 'I' * num
	elif num < 9:
		return self.intToRoman(5) + self.intToRoman(num - 5)
	elif num < 40:
		return 'X' * (num // 10) + self.intToRoman(num % 10)
	elif num < 50:
		return self.intToRoman(40) + self.intToRoman(num - 40)
	elif num < 90:
		return self.intToRoman(50) + self.intToRoman(num - 50)
	elif num < 100:
		return self.intToRoman(90) + self.intToRoman(num % 90)
	elif num < 400:
		return 'C' * (num // 100) + self.intToRoman(num % 100)
	elif num < 500:
		return self.intToRoman(400) + self.intToRoman(num - 400)
	elif num < 900:
		return self.intToRoman(500) + self.intToRoman(num - 500)
	elif num < 1000:
		return self.intToRoman(900) + self.intToRoman(num % 900)
	else:
		return 'M' * (num // 1000) + self.intToRoman(num % 1000)

算法分析：

时间复杂度为O(1)，空间复杂度O(1)