LeetCode-012整数转罗马数字

坚持真的很难 因为有事断了一星期

public String intToRoman(int num) {

    if(num>3999||num<1){//不在范围内直接return
        return null;
    }

    //规则存在HashMap方便取
    HashMap<Integer,String> hashMap = new HashMap<Integer,String>(){
        {
            put(1,"I");
            put(5,"V");
            put(10,"X");
            put(50,"L");
            put(100,"C");
            put(500,"D");
            put(1000,"M");
            put(4,"IV");
            put(9,"IX");
            put(40,"XL");
            put(90,"XC");
            put(400,"CD");
            put(900,"CM");
        }
    };

    String res = "";//结果
    int i = 1;//所在位置 个位 十位 百位。。。

    //从个位开始取
    while(num>0){
        int temp = num % 10;//当前位置数值
        int key = temp * i;//其实际表示数值
        int center = key;//678大于5逐步减到5中断
        if(hashMap.containsKey(key)){//在规则种存在直接连接
            res = hashMap.get(key) + res;
        }else {
            for(int j=0;j<temp;j++){//在规则中不存在，向下找规则连接
                res = hashMap.get(key/temp) + res;
                center = center - key/temp;//678大于5的数字逐步减到5
                if(hashMap.containsKey(center)){//把5规则连接，跳出for循环
                    res = hashMap.get(center) + res;
                    break;
                }
            }
        }
        i = i * 10;
        num = num / 10;
    }

    return res;

}