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"};

580ms+，时间有点长

 

仔细观察罗马数字的话，可以把每个字母直接由一个数值替代，然后将他们累加，不过有个例外就是遇到4,9时要看字母前面一个字母的数值比如IV如果小于后面的则实际对应值为V-I = 5-1 = 4，X - I= 10 - 1 = 9

具体可以参考：http://www.cnblogs.com/zhuli19901106/p/3453180.html