27-Roman to Integer-Leetcode

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Given a roman numeral, convert it to an integer.

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

思路：先构造罗马与数字的映射，接着对给定的字符串从头开始进行最长匹配，最大为4，依次进行，匹配完成则乘以相应位的量级，时间复杂度为O(n)

#include <algorithm>#include <iostream>#include <vector>#include <string>#include <map>#define IMIN numeric_limits<int>::min()#define IMAX numeric_limits<int>::max()#define FR(i,n) for(int i=0;i<n;i++)#define CLC(x) memset(x,0,sizeof(x))#define FILL(x,c) memset(x,c,sizeof(x))using namespace std;class Solution {public:    int romanToInt(string s) {        map<string,int> mp[4];        init4(mp[3]);init3(mp[2]);init2(mp[1]);init1(mp[0]);        int n=s.size();        int res=0;        for(int i=0;i<n;++i)        {            int len = min(n-i,4);            int go_len = 0;            while(len)            {                string tmp(s,i,len);                int flag =0;                for(int j=3;j>=0;--j)                    if(mp[j].count(tmp)){                        go_len = len;                         res+=pow(10,j)*mp[j][tmp];                        flag = 1; break;                    }                if(flag==1)break;                len--;            }            i = i+go_len-1;        }        cout<<res<<endl;        return res;    }    void init1(map<string,int> &mp)    {        mp[""]=0;        mp[string("I")]=1;mp[string("II")]=2;        mp[string("III")]=3;mp[string("IV")]=4;        mp[string("V")]=5;mp[string("VI")]=6;        mp[string("VII")]=7;mp[string("VIII")]=8;        mp[string("IX")]=9;         }    void init2(map<string,int> &mp)    {        mp[""]=0;        mp[string("X")]=1;mp[string("XX")]=2;        mp[string("XXX")]=3;mp[string("XL")]=4;        mp[string("L")]=5;mp[string("LX")]=6;        mp[string("LXX")]=7;mp[string("LXXX")]=8;        mp[string("XC")]=9;         }    void init3(map<string,int> &mp)    {        mp[""]=0;        mp[string("C")]=1;mp[string("CC")]=2;        mp[string("CCC")]=3;mp[string("CD")]=4;        mp[string("D")]=5;mp[string("DC")]=6;        mp[string("DCC")]=7;mp[string("DCCC")]=8;        mp[string("CM")]=9;         }    void init4(map<string,int> &mp)    {        mp[string("M")]=1;mp[string("MM")]=2;        mp[string("MMM")]=3;    }};int main(){    Solution s;    string str;    while(cin>>str)    cout<<s.romanToInt(str)<<endl;    return 0;}

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