题目：原题链接（中等）
标签：字符串、数学、贪心算法
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O(1) | O(1) | 60ms (72.76%) |
Ans 2 (Python) | O(1) | O(1) | 56ms (85.08%) |
Ans 3 (Python) | O(1) | O(1) | 68ms (39.91%) |
LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。
解法一（硬编码数字法）：
def intToRoman(self, num: int) -> str:
def translate(n, sign1, sign2, sign3):
if n == 0:
return ""
elif 1 <= n <= 3:
return sign1 * n
elif n == 4:
return sign1 + sign2
elif 5 <= n <= 8:
return sign2 + sign1 * (n - 5)
elif n == 9:
return sign1 + sign3
a = translate(num // 1000, "M", "", "")
b = translate(num % 1000 // 100, "C", "D", "M")
c = translate(num % 100 // 10, "X", "L", "C")
d = translate(num % 10, "I", "V", "X")
return a + b + c + d解法二（更彻底的硬编码数字法）：
def intToRoman(self, num: int) -> str:
B4 = ["", "M", "MM", "MMM"]
B3 = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"]
B2 = ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"]
B1 = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"]
return B4[num // 1000] + B3[num % 1000 // 100] + B2[num % 100 // 10] + B1[num % 10]解法三（贪心算法）：
def intToRoman(self, num: int) -> str:
table = [(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")]
ans = []
for n, s in table:
count, num = divmod(num, n)
ans.append(count * s)
return "".join(ans)
