【LeetCode】13. Roman to Integer (2 solutions)

Roman to Integer

Given a roman numeral, convert it to an integer.

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

解法一：非递归

从左到右遍历每个字符，并记录上个字符来处理双字符情况即可。

class Solution {
public:
    int romanToInt(string s) {
        int n = ;
        char lastC = ;
        for(int i = ; i < s.size(); i ++)
        {
            switch(s[i])
            {
                case 'I':
                    n += ;
                    lastC = s[i];
                    break;
                case 'V':
                    if(lastC == 'I')
                    {//IV
                        n -= ;
                        n += ;
                        lastC = ;
                    }
                    else
                    {
                        n += ;
                        lastC = s[i];
                    }
                    break;
                case 'X':
                    if(lastC == 'I')
                    {//IX
                        n -= ;
                        n += ;
                        lastC = ;
                    }
                    else
                    {
                        n += ;
                        lastC = s[i];
                    }
                    break;
                case 'L':
                    if(lastC == 'X')
                    {//XL
                        n -= ;
                        n += ;
                        lastC = ;
                    }
                    else
                    {
                        n += ;
                        lastC = s[i];
                    }
                    break;
                case 'C':
                    if(lastC == 'X')
                    {//XC
                        n -= ;
                        n += ;
                        lastC = ;
                    }
                    else
                    {
                        n += ;
                        lastC = s[i];
                    }
                    break;
                case 'D':
                    if(lastC == 'C')
                    {//CD
                        n -= ;
                        n += ;
                        lastC = ;
                    }
                    else
                    {
                        n += ;
                        lastC = s[i];
                    }
                    break;
                case 'M':
                    if(lastC == 'C')
                    {//CM
                        n -= ;
                        n += ;
                        lastC = ;
                    }
                    else
                    {
                        n += ;
                        lastC = s[i];
                    }
                    break;
                default:
                    return ;
            }
        }
        return n;
    }
};

解法二：递归

处理开头字符情况之后递归处理后续字符串

class Solution {
public:
    int romanToInt(string s) {
        if(s == "")
            return ;
        //to here, s.size() >= 1
        else if(s[] == 'M')
            return  + romanToInt(s.substr());
        else if(s[] == 'C')
        {
            if(s.size() == )
            //s == "C"
                return ;
            if(s[] == 'M')
                return  + romanToInt(s.substr());
            else if(s[] == 'D')
                return  + romanToInt(s.substr());
            else
                return  + romanToInt(s.substr());
        }
        else if(s[] == 'D')
            return  + romanToInt(s.substr());
        else if(s[] == 'X')
        {
            if(s.size() == )
            //s == "X"
                return ;
            if(s[] == 'C')
                return  + romanToInt(s.substr());
            else if(s[] == 'L')
                return  + romanToInt(s.substr());
            else
                return  + romanToInt(s.substr());
        }
        else if(s[] == 'L')
            return  + romanToInt(s.substr());
        else if(s[] == 'I')
        {
            if(s.size() == )
            //s == "I"
                return ;
            if(s[] == 'X')
                return  + romanToInt(s.substr());
            else if(s[] == 'V')
                return  + romanToInt(s.substr());
            else
                return  + romanToInt(s.substr());
        }
        else if(s[] == 'V')
            return  + romanToInt(s.substr());
    }
};