Roman to Integer

Given a roman numeral, convert it to an integer.

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

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按照intger to roman 的思路，写下面这道题，比较绕， 而且写的不够简洁，而且修改了好几次

class Solution {
public:
    int romanToInt(string s) {
        string c[4][10]={{"0","I","II","III","IV","V","VI","VII","VIII","IX"},
                         {"0","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
                         {"0","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
                         {"0","M","MM","MMM"}};

        int len = s.length();
        if (0 == len) {
            return 0;
        }
        int begin = 0;
        int i;
        int res = 0;
        while (begin < len) {
            if (s[begin] == 'M') {
                for (i=3; i>0; i--) {
                    if (s.find(c[3][i]) != string::npos) {
                        res += i * 1000;
                        begin = begin + c[3][i].length();
                        break;
                    }
                }
            } else if (s[begin] == 'C' || s[begin] == 'D') {
                for (i=9; i>0; i--) {
                    if (s.find(c[2][i]) != string::npos) {
                        //当找到D的时候，并不能够确定是找到CD 还是D
                        //做一下判断
                        if (begin >=0 && s[begin] == 'C' && s[begin + 1] == 'D') {
                            res += 4 * 100;
                            begin = begin + c[2][4].length();
                            break;
                        } else {
                            res += i * 100;
                            begin = begin + c[2][i].length();
                            break;
                        }
                    }
                }
            } else if (s[begin] == 'X' || s[begin] == 'L') {
                for (i=9; i>0; i--) {
                    if (s.find(c[1][i]) != string::npos) {
                        if (begin >=0 && s[begin] == 'X' && s[begin + 1] == 'L') {
                            res += 4 * 10;
                            begin = begin + c[1][4].length();
                            break;
                        } else {
                            res += i * 10;
                            begin = begin + c[1][i].length();
                            break;
                        }
                    }
                }
            } else if (s[begin] == 'I' || s[begin] == 'V') {
                for (i=9; i>0; i--) {
                    if (s.find(c[0][i]) != string::npos) {
                        if (begin >=0 && s[begin] == 'I' && s[begin + 1] == 'V') {
                            res += 4 ;
                            begin = begin + c[0][4].length();
                            break;
                        } else {
                            res += i;
                            begin = begin + c[0][i].length();
                            break;
                        }
                    }
                }
            } else {
                return res;
            }
        }
        return res;
    }
};

如果按照下面这种思路写就很简单了

class Solution {
public:
    int romanToInt(string s) {
        map<char,int> roman;
        roman['I']=1;
        roman['V']=5;
        roman['X']=10;
        roman['L']=50;
        roman['C']=100;
        roman['D']=500;
        roman['M']=1000;

        int len = s.length();
        if (0 == len) {
            return 0;
        }

        int res = 0;
        int index;
        for (index=0; index<len; index++) {
            if (index>0 && roman[s[index]] > roman[s[index -1]]) {
                res = res + (roman[s[index]] -2 * roman[s[index -1]]);
            } else {
                res = res + roman[s[index]];
            }
        }

        return res;
    }

};