题意：给你一个数字，转换成罗马数字，需要遵循罗马数字排布的一些规则。
解题思路：
思路一：因为9和4比较特殊，其余的就是直接累计，可以把罗马数字分成9,5,4,1的对应表，然后逐一遍历将对应的罗马符号放入该位即可。

class Solution {
public:
    string intToRoman(int num) {
        string ans;
        
        vector<vector<string>> ivec;      
        vector<string> ivec1 = {"IX","V","IV","I"};
        vector<string> ivec2 = {"XC","L","XL","X"};
        vector<string> ivec3 = {"CM","D","CD","C"};
        vector<string> ivec4 = {"  "," ","  ","M"};
        ivec.push_back(ivec1);
        ivec.push_back(ivec2);
        ivec.push_back(ivec3);
        ivec.push_back(ivec4);
        
        int a[4] = {9, 5, 4, 1};
        int bt = 0;
        while(num){
            int n = num % 10;
            string ss;
            while(n){
                for(int i = 0; i < 4; i++){
                    //cout << a[i]  << endl;
                    if(n >= a[i]){
                        ss += ivec[bt][i];
                        n -= a[i];
                    }
                }
            }
            ans = ss + ans;
            num /= 10;
            bt++;
        }
        return ans;
    }
};

思路二：由于题目给出条件，说数字不会超过3999，直接暴力更快更准确。

class Solution {
public:
    string intToRoman(int num) {
        string a[10] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
        string b[10] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
        string c[10] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
        string d[10] = {"", "M", "MM", "MMM", "", "", "", "", "", ""};
        return d[num/1000] + c[(num/100)%10] + b[(num/10)%10] + a[num%10]; 
    }
};

真 * 牛逼！