leetcode 13

Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.

// 1~9: {“I”, “II”, “III”, “IV”, “V”, “VI”, “VII”, “VIII”, “IX”};
// 10~90: {“X”, “XX”, “XXX”, “XL”, “L”, “LX”, “LXX”, “LXXX”, “XC”};
// 100~900: {“C”, “CC”, “CCC”, “CD”, “D”, “DC”, “DCC”, “DCCC”, “CM”};
// 1000~3000: {“M”, “MM”, “MMM”}.

class Solution {public:    int romanToInt(string s) {        map<char, int> r;        int ret = 0;;        r['I'] = 1;        r['V'] = 5;        r['X'] = 10;        r['L'] = 50;        r['C'] = 100;        r['D'] = 500;        r['M'] = 1000; // r.insert(make_pair('M', 1000));        typedef string::size_type sz;        int nlast = 0;        int nthis = 0;        for (sz i = 0; i != s.size(); ++i) {            if (r.find(s[i]) == r.end()) return 0;            nthis = r[s[i]];            if (nthis > nlast)                ret += nthis - nlast -nlast;            else ret += nthis;            nlast = nthis;        }        return ret;    }};

参考后

class Solution {public:    int romanToInt(string s) {        map<char, int> r = {{'I', 1}, {'V', 5},                             {'X', 10}, {'L', 50},                            {'C', 100}, {'D', 500},                            {'M', 1000}};        int ret = 0;        int nlast = 0;        int nthis = 0;        for (string::size_type i = s.size() - 1; i != -1; --i) {            if (r.find(s[i]) == r.end()) return 0;            nthis = r[s[i]];            if (nthis < nlast)                ret -= nthis;            else ret += nthis;            nlast = nthis;        }        return ret;    }};