写法1

class Solution {
public:
    int romanToInt(string s) {
    int n=s.size(),ans=0;
    unordered_map<char,int> words;
    words['I']=1,words['V']=5;
    words['X']=10,words['L']=50;
    words['C']=100,words['D']=500;
    words['M']=1000;
    for(int i=0;i<n;i++)
    {
        if(i!=n-1&&words[s[i+1]]>words[s[i]])
        {
           ans+=words[s[i+1]]-words[s[i]];
           i++;
        }
        else 
        ans+=words[s[i]];
    }
    return ans;
    }

};

写法二

class Solution {
public:
    int romanToInt(string s) {
    s+="I";//便于for循环不用判断溢出
    int n=s.size(),ans=0;
    unordered_map<char,int> words;
    words['I']=1,words['V']=5;
    words['X']=10,words['L']=50;
    words['C']=100,words['D']=500;
    words['M']=1000;
    for(int i=0;i<n-1;i++)
    {
        if(words[s[i]]>=words[s[i+1]])//i+1不会溢出
        ans+=words[s[i]];
        else 
        ans-=words[s[i]];
    }
    return ans;
    }

};