Leetcode: Roman to Integer

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

利用上一题的程序：

Judge Small： Accepted!

Judge Large：Time Limit Exceeded

map<string,int> romanmp;
    void romanstr(string& roman, int num, char* symbol)
	{
		if(num == 0)return;
		else if(num <= 3) roman.append(num, *symbol);
		else if(num == 4){
			roman.append(1,*symbol);
			roman.append(1,*(symbol+1));
		}else if(num <= 8){
			roman.append(1,*(symbol+1));
			roman.append(num-5,*symbol);
		}else if(num == 9){
			roman.append(1,*symbol);
			roman.append(1,*(symbol+2));
		}
	}
	string intToRoman(int num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        char symbol[9] = {'I','V','X','L','C','D','M'};
        string roman="";
		int scale = 1000;
		int p=6;
		while(num)
		{
			int bit = num/scale;
			romanstr(roman, bit, symbol+p);
			num = num%scale;
			scale /= 10;
			p -= 2;
		}
		return roman;
    }
	int romanToInt(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        for(int i = 0;i <4000; i++)
			romanmp[intToRoman(i)] = i;
		return romanmp[s];
    }

Accepted!

int romanToInt(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int length = s.length();
    	if(length <1) return 0;
		map<char,int> m;
		m['I'] = 1;
		m['V'] = 5;
		m['X'] = 10;
		m['L'] = 50;
		m['C'] = 100;
		m['D'] = 500;
		m['M'] = 1000;
		int i = length-1;
		int sum = m[s[i--]];
		while(i>=0)
			if(m[s[i+1]] > m[s[i]])
				sum -= m[s[i--]];
			else
				sum += m[s[i--]];
		return sum;
    }