[Leedcode]Roman to Integer

最新推荐文章于 2021-10-14 10:46:05 发布

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本文介绍了一种将罗马数字转换为整数的方法，通过解析罗马数字的构成规则，实现从1到3999范围内罗马数字的有效转换。文章详细解释了罗马数字的书写规则，包括特殊情况下的减法规则，并提供了一个Python实现示例。

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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 a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

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

Example 5:

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

代码：

class Solution(object):
def romanToInt(self, s):
"""
:type s: str
:rtype: int
"""
key_sym = ['I', 'V', 'X', 'L', 'C', 'D', 'M']
key_val = [1, 5, 10, 50, 100, 500, 1000]

# create dict
key_dict = {}
for i in range(len(key_sym)):
key_dict[key_sym[i]] = key_val[i]

s_list = list(s)

nums = 0
while s_list:
cur_keys = []
sym_f = s_list[0]
ind_f = key_sym.index(sym_f)

# the last symbol
if len(s_list) >1:
j = 1
sym_l = s_list[j]
ind_l = key_sym.index(sym_l)
else:
ind_l = ind_f

if ind_f >= ind_l:
nums += key_dict[sym_f]
s_list.pop(0)
elif ind_f < ind_l:
nums = nums + key_dict[sym_l] - key_dict[sym_f]
s_list.pop(0)
s_list.pop(0)
return nums