leetcode 12 整数转罗马数字 贪心

额，连着两个贪心？

这是局部最优问题：能用大“罗马数表示”就不会用小的。

先构造出所有基础罗马数，然后从大到小比较

因为比较的只有1000，900，...有限并有些麻烦，构造table  map<int,string>

然后，map默认安装按照key的值升序排序..

想从大到小，用reverse_iterator

class Solution {
public:
    string intToRoman(int num) {
         map<int,string> calc = {{1000,"M"},{900,"CM"},{500,"D"},{400,"CD"},{100,"C"}, {90,"XC"},{50,"L"},{40,"XL"},{10,"X"},{9,"IX"},{5,"V"},{4,"IV"},{1,"I"}};
        map<int,string>::reverse_iterator iter=calc.rbegin();
        string ret;
        while(iter!=calc.rend())
        {
            if(num >= iter->first)
            {
                ret += iter->second;
                num-= iter->first;
            }
            else
                iter++;
        }
        return ret;
    }
};

原文地址：https://www.cnblogs.com/lqerio/p/11750057.html