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.
题意:题目要求将罗马数字转换成为整数。具体的转换规则其实题目也已经给出,根据给出规则,判断每个字符累加即可
C++:
class Solution {
public:
int romanToInt(string s) {
unordered_map<char,int> m{{'I',1}, {'V',5}, {'X',10}, {'L',50}, {'C',100}, {'D',500}, {'M',1000}};
int res = 0;
for(int i = 0; i < s.length(); i++){
int val = m[s[i]];
if(i == s.length()-1 || m[s[i+1]] <= m[s[i]])res += val;
else
res -= val;
}
return res;
}
};
Python3:
class Solution:
def romanToInt(self, s: str) -> int:
m = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
res = 0
len_s = len(s)
for i in range(len_s):
if i == len_s-1 or m[s[i+1]] <= m[s[i]]:
res += m[s[i]]
else:
res -= m[s[i]]
return res