1、题目
13. Roman to Integer
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 beforeV
(5) andX
(10) to make 4 and 9. -
X
can be placed beforeL
(50) andC
(100) to make 40 and 90. -
C
can be placed beforeD
(500) andM
(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.
2、我的解法
罗马数字转化成阿拉伯数字,多重判断。。。。
# -*- coding: utf-8 -*-
# @Time : 2020/1/28 14:35
# @Author : SmartCat0929
# @Email : 1027699719@qq.com
# @Link : https://github.com/SmartCat0929
# @Site :
# @File : 13. Roman to Integer(字符串索引).py class Solution:
def romanToInt(self, s: str) -> int:
l = len(s)
y = 0
for i in range(l):
if s[i]=="V"and s[i-1]=="I"and i!=0:
y=y+3
elif s[i]=="X"and s[i-1]=="I"and i!=0:
y=y+8
elif s[i]=="L"and s[i-1]=="X"and i!=0:
y=y+30
elif s[i]=="C"and s[i-1]=="X"and i!=0:
y=y+80
elif s[i]=="D"and s[i-1]=="C"and i!=0:
y=y+300
elif s[i]=="M"and s[i-1]=="C"and i!=0:
y=y+800
elif s[i] == "I":
y = y + 1
elif s[i] == "V":
y = y + 5
elif s[i] == "X":
y = y + 10
elif s[i]== "L":
y = y + 50
elif s[i] == "C":
y = y + 100
elif s[i] == "D":
y = y + 500
elif s[i] == "M":
y = y + 1000
if y>=1 and y<=3999:
return y
else:
return 0 print (Solution().romanToInt("MMMCDXC"))