13. Roman to Integer (Easy)
https://leetcode.com/problems/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.
Solutions
1.
class Solution {
public int romanToInt(String s) {
int ans = 0;
if (s == null || s.isEmpty()) {
return ans;
}
// Build the map between the symbols and corresponding values
int[] nums = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
String[] roman = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"};
while (!s.isEmpty()) {
String rn = "";
for (int i = roman.length - 1; i >= 0; i--) {
if (s.startsWith(roman[i])) {
rn = roman[i];
ans += nums[i];
break;
}
}
// rn can not be empty
s = s.substring(rn.length());
}
return ans;
}
}
2.
class Solution {
public int romanToInt(String s) {
int ans = 0;
if (s == null || s.isEmpty()) {
return ans;
}
// Build the map between the symbols and corresponding values
Map<Character, Integer> map = new HashMap<>();
map.put('I', 1);
map.put('V', 5);
map.put('X', 10);
map.put('L', 50);
map.put('C', 100);
map.put('D', 500);
map.put('M', 1000);
int pre = 0;
// XLVI=46, L=50, X=10, V=5, I=1
for (int i = 0; i < s.length(); i++) {
Character c = s.charAt(i);
int val = map.get(c);
// set the previous value for the reason that there is no previous value at first position.
if (i == 0) {
pre = val;
continue;
}
// minus it if previous value is less than current, such as XL, X is -10
if (pre < val) {
ans -= pre;
}
// plus it once previous value is greater than current, for example LX, X is 10
else {
ans += pre;
}
pre = val;
}
// The last position value is not added. The fact is that the last value is always a positive one.
return ans + pre;
}
}