Leetcode Roman to integer

class Solution {private:    const static char* pattern[];    const static char* roman[];    unordered_map<string, int> a2i;public:    int romanToInt(string s) {        int res = 0;        if (a2i.size() == 0) build_tlb();                while (!s.empty()) {            int len = s.length();            for (int i=min(4, len); i>0; i--) {                unordered_map<string, int>::iterator iter = a2i.find(s.substr(len-i));                if (iter == a2i.end()) continue;                res += iter->second;                s.resize(len - i);                break;            }        }                return res;    }        void build_tlb() {        int pw = 1;        for (int i=0; i<3; i++) {            for (int j=1; j<=9; j++) {                string rm;                for (int k=0; pattern[j][k] != ‘\0‘; k++) {                    rm.push_back(roman[i][ pattern[j][k] - ‘0‘ ]);                }                a2i.insert(make_pair(rm, pw * j));            }            pw *= 10;        }            a2i.insert(make_pair("M", 1000));        a2i.insert(make_pair("MM", 2000));        a2i.insert(make_pair("MMMM", 3000));    }};const char* Solution::pattern[] = {"A", "0", "00", "000", "01", "1", "10", "100", "1000", "02"};const char* Solution::roman[] = {"IVX", "XLC", "CDM", "M"};

580 MS +, long time

 

If you carefully observe the roman numerals, you can replace each letter with a numerical value and then accumulate them, but there is an exception is to encounter 4, 9 to see a letter before the value of a letter, such as IV if less than the following, the actual corresponding value is V-I = 5-1 = 4, x-I = 10-1 = 9

Specific can refer to: http://www.cnblogs.com/zhuli19901106/p/3453180.html