Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
| public class Solution {
public int romanToInt(String s) {
if (s == null || s.length() < 1){return 0;}
int[] values = new int[26];
values['I' - 'A'] = 1;
values['V' - 'A'] = 5;
values['X' - 'A'] = 10;
values['L' - 'A'] = 50;
values['C' - 'A'] = 100;
values['D' - 'A'] = 500;
values['M' - 'A'] = 1000;
char[] chars = s.toCharArray();
int res = 0, prev = 1000;
for (char c : chars){
int cur = values[c - 'A'];
if (cur > prev){
if (cur / prev > 10){return 0;}
res -= 2 * prev;
}
res += cur;
prev = cur;
}
return res;
}
}
|
The first solution is much faster than using a HashMap
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
| public class Solution {
public int romanToInt(String s) {
if (s == null || s.length() < 1){return 0;}
HashMap<Character, Integer> values = new HashMap<Character, Integer>();
values.put('I',1);
values.put('V',5);
values.put('X',10);
values.put('L',50);
values.put('C',100);
values.put('D',500);
values.put('M',1000);
char[] chars = s.toCharArray();
int res = 0, prev = 1000;
for (char c : chars){
int cur = values.get(c);
if (cur > prev){
if (cur / prev > 10){return 0;}
res -= 2 * prev;
}
res += cur;
prev = cur;
}
return res;
}
}
|