Roman numerals are represented by seven different symbols: `I`

, `V`

, `X`

, `L`

, `C`

, `D`

and `M`

.

SymbolValueI 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 one's 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 before`V`

(5) and`X`

(10) to make 4 and 9.`X`

can be placed before`L`

(50) and`C`

(100) to make 40 and 90.`C`

can be placed before`D`

(500) and`M`

(1000) to make 400 and 900.

Given a roman numeral, convert it to an integer.

**Example 1:**

Input:s = "III"Output:3

**Example 2:**

Input:s = "IV"Output:4

**Example 3:**

Input:s = "IX"Output:9

**Example 4:**

Input:s = "LVIII"Output:58Explanation:L = 50, V= 5, III = 3.

**Example 5:**

Input:s = "MCMXCIV"Output:1994Explanation: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]`

.

class Solution {
public int romanToInt(String s) {
if(s==null){
return 0;
}
int len = s.length();
if(len==0){
return 0;
}
Map<Character, Integer> map = new HashMap<Character, Integer>();
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 num = map.get(s.charAt(len-1));
for(int i=len-2;i>=0;i--){
int currVal = map.get(s.charAt(i));
int nextVal = map.get(s.charAt(i+1));
if(currVal < nextVal){
num -= currVal;
}
else{
num += currVal;
}
}
return num;
}
}

This question requires the knowledge of Roman Numbers, If we know how we can use the additive and subtractive nature of Roman numbers then this question is very simple.

Time Complexity: O(n) //Where n is the length of Roman String

Space Complexity: O(1) //No extra space is used.