Leetcode 0013. 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, `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`.
Description
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
- III = 3.
Example 2:
Input: s = "LVIII" Output: 58
- L = 50, V= 5, III = 3.
Example 3:
Input: s = "MCMXCIV" Output: 1994
- 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]
.
Solution
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class Solution {
/**
* Iteration
* Time Complexity: BigO(N)
* Space Complexity: BigO(N)
*/
public int romanToInt(String s) {
int result = 0, num = 0;
for (int i = s.length()-1; i >= 0; i--) {
switch(s.charAt(i)) {
case 'I':
num = 1;
break;
case 'V':
num = 5;
break;
case 'X':
num = 10;
break;
case 'L':
num = 50;
break;
case 'C':
num = 100;
break;
case 'D':
num = 500;
break;
case 'M':
num = 1000;
break;
}
if (4*num < result) result -= num;
else result += num;
}
return result;
}
}
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