13. Roman to Integer
Problem
Tags: Hash Table
, Math
, String
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 ones 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 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.
Example 1:
Input: s = "III"
Output: 3
Explanation: III = 3.
Example 2:
Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 3:
Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
Constraints:
1 <= s.length <= 15
s
contains only the characters('I', 'V', 'X', 'L', 'C', 'D', 'M')
.- It is guaranteed that
s
is a valid roman numeral in the range[1, 3999]
.
Code
C
// 13. Roman to Integer (1/27/54192)
// Runtime: 4 ms (85.70%) Memory: 5.97 MB (22.73%)
int romanToInt (char* s) {
int dict[128] = { 0 };
dict['I'] = 1;
dict['V'] = 5;
dict['X'] = 10;
dict['L'] = 50;
dict['C'] = 100;
dict['D'] = 500;
dict['M'] = 1000;
int len = strlen(s);
int ans = dict[s[len - 1]];
for (int i = 0; i < len - 1; i++) {
if (dict[s[i]] >= dict[s[i + 1]]) {
ans += dict[s[i]];
} else {
ans -= dict[s[i]];
}
}
return ans;
}
TS
// 13. Roman to Integer (1/21/54192)
// Runtime: 230 ms (24.32%) Memory: 48.64 MB (42.38%)
function romanToInt(s: string): number {
const dict = { M: 1000, D: 500, C: 100, L: 50, X: 10, V: 5, I: 1 };
let ans = dict[s[s.length - 1]];
for (let i = 0; i < s.length - 1; i++) {
if (dict[s[i]] < dict[s[i + 1]]) {
ans -= dict[s[i]];
} else {
ans += dict[s[i]];
}
}
return ans;
};