# C solution to Roman2Int problem

• It is obvious that Roman numerals consist of 'I', 'V', 'X', 'L', 'C', 'D', 'M', representing for 1, 5, 10, 50, 100, 500, 1000. We also know that a number reside to the left of a larger number means subtraction operation. Symmetrically, a number reside to the right of an equal or larger number means addition operation. With that in mind, let's look at 2 examples:

1. IV
The answer is 4, because 'I' is smaller than as well as left to 'V', so we interpret it as: 'V' - 'I' = 4.
2. LXV
The answer is 65, because 'X' is larger than and right to 'V', so we add them: 'X' + 'V' = 15; meanwhile, 'L' and 'X' show us the same pattern, we we do addition again: 'L' + previous result = 65.

Now we have no trouble interpreting Roman numerals. In my solution, I firstly define a function for the convenience of transferring a Roman char to an int type, then apply algorithm on it:

``````int charToInt( char c ) {
switch( c ) {
case 'I':
return 1;
case 'V':
return 5;
case 'X':
return 10;
case 'L':
return 50;
case 'C':
return 100;
case 'D':
return 500;
case 'M':
return 1000;
default:
return -1;
}
}

int romanToInt(char* s) {
char prev, curr;
int sum, len, a, b;
len = strlen( s );
prev = s[len - 1];
sum = charToInt( prev );

while( --len > 0 ) {
curr = s[len - 1];
a = charToInt( prev );
b = charToInt( curr ) ;
prev = curr;
sum = a > b ? sum - b : sum + b;
}

return sum;
}
``````

However, my solution does not check error for illegal inputs that violate the rules of representing Roman numerals. Solution rests on the surmise of correct input.

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