[数据结构与算法]LeetCode 12. 整数转罗马数字

难度：中等

解题思路：和我之前做的罗马转整数一样，暴力就完事了，个人感觉难度应该是简单，上代码

class Solution {
    public String intToRoman(int num) {
         HashMap<Integer,String> map = new HashMap<>();
        map.put(1,"I");
        map.put(4,"IV");
        map.put(5,"V");
        map.put(9,"IX");
        map.put(10,"X");
        map.put(40,"XL");
        map.put(50,"L");
        map.put(90,"XC");
        map.put(100,"C");
        map.put(400,"CD");
        map.put(500,"D");
        map.put(900,"CM");
        map.put(1000,"M");
        String result = "";
        while (num!=0){
            if (num >=1000){
                num = num-1000;
                result += map.get(1000);
            }else if (num >= 900){
                num = num-900;
                result += map.get(900);
            }else if (num >= 500){
                num = num-500;
                result += map.get(500);
            }else if (num >= 400){
                num = num-400;
                result += map.get(400);
            }else if (num >= 100){
                num = num-100;
                result += map.get(100);
            }else if (num >= 90){
                num = num-90;
                result += map.get(90);
            }else if (num >= 50){
                num = num-50;
                result += map.get(50);
            }else if (num >= 40){
                num = num-40;
                result += map.get(40);
            }else if (num >= 10){
                num = num-10;
                result += map.get(10);
            }else if (num >= 9){
                num = num-9;
                result += map.get(9);
            }else if (num >= 5){
                num = num-5;
                result += map.get(5);
            }else if (num >= 4){
                num = num-4;
                result += map.get(4);
            }else if (num >= 1){
                num = num-1;
                result += map.get(1);
            }
        }
        return result;
    }
}