LeetCode13. 罗马数字转整数

题目

分析

受LeetCode 12题目影响，一时没反应过来用哈希存储，虽说难度是简单，还没上一道做的舒服。。。。

这道题目关键在于特判下4的特殊情况，就是后一个字母比当前字母要大时情况要特殊判断下

代码

class Solution {
public:
    int romanToInt(string s) {

        unordered_map<char,int>m;
        m['M'] = 1000;
        m['D'] = 500;
        m['C'] = 100;
        m['L'] = 50;
        m['X'] = 10;
        m['V'] = 5;
        m['I'] = 1;

        int res  = 0;
        for(int i = 0; i < s.size();i++){
            //cout<<s[i] << " "<<s[i+1] << " "<<m[s[i]] << " "<< m[s[i+1]]<<endl;
            if(m[s[i]] < m[s[i+1]] && i < s.size() -1){
                res += (m[s[i+1]] - m[s[i]]);
                i = i + 1;
            }else{
                res += m[s[i]];
            }
            //cout<<res<<endl;
            
        }
        return res;
    }
};

 

y总：

class Solution {
public:
    int romanToInt(string s) {

        unordered_map<char,int>m;
        m['M'] = 1000;
        m['D'] = 500;
        m['C'] = 100;
        m['L'] = 50;
        m['X'] = 10;
        m['V'] = 5;
        m['I'] = 1;

        int res  = 0;
        for(int i = 0; i < s.size();i++){
            if(m[s[i]] < m[s[i+1]] && i < s.size() -1){
                res -= m[s[i]];
            }else{
                res += m[s[i]];
            }
            
        }
        return res;
    }
};