LeetCode 12.Integer to Roman Arabic numerals converted to Roman numerals

LeetCode 12.Integer to Roman Arabic numerals converted to Roman numerals

12. Integer to Roman

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9.
• X can be placed before L (50) and C (100) to make 40 and 90.
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: "III"

Example 2:

Input: 4
Output: "IV"

Example 3:

Input: 9
Output: "IX"

Example 4:

Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.

Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Recursion

By comparing and Roman numeral converter. But recursion is inefficient, many repetitive calculations.

class Solution {
    public String intToRoman(int num) {
        return num >= 1000 ? "M" + intToRoman(num-1000) : 
        num >= 900 ? "CM" + intToRoman(num-900) :
        num >= 500 ? "D" + intToRoman(num-500) :
        num >= 400 ? "CD" + intToRoman(num-400) :
        num >= 100 ? "C" + intToRoman(num-100) :
        num >= 90 ? "XC" + intToRoman(num-90) :
        num >= 50 ? "L" + intToRoman(num-50) :
        num >= 40 ? "XL" + intToRoman(num-40) :
        num >= 10 ? "X" + intToRoman(num-10) :
        num >= 9 ? "IX" + intToRoman(num-9) :
        num >= 5 ? "V" + intToRoman(num-5) :
        num >= 4 ? "IV" + intToRoman(num-4) :
        num >= 1 ? "I" + intToRoman(num-1) : "";
    }
}

Direct conversion

Recursion is very low efficiency, we can according to the conversion rules using an iterative fashion. Each rule is seen as a key-value pair. Establish a RomanMap object. According to the size and number of key-value pairs of Arabic numerals and descending to compare, and then transformed. There is provided a variable lastPosition order to reduce the alignment. Although the establishment of an object by way of efficiency will be a little low, but will increase the readability of the code.

class Solution {
    class RomanMap{
        int number;
        String romanNumber;
        
        RomanMap(int number,String romanNumber){
            this.number = number;
            this.romanNumber = romanNumber;
        }
        
        int getNumber(){
            return number;
        }
        
        String getRomanNumber(){
            return romanNumber;
        }
    }
    public String intToRoman(int num) {
        List<RomanMap> list = new ArrayList<RomanMap>();
        list.add(new RomanMap(1000,"M"));
        list.add(new RomanMap(900,"CM"));
        list.add(new RomanMap(500,"D"));
        list.add(new RomanMap(400,"CD"));
        list.add(new RomanMap(100,"C"));
        list.add(new RomanMap(90,"XC"));
        list.add(new RomanMap(50,"L"));
        list.add(new RomanMap(40,"XL"));
        list.add(new RomanMap(10,"X"));
        list.add(new RomanMap(9,"IX"));
        list.add(new RomanMap(5,"V"));
        list.add(new RomanMap(4,"IV"));
        list.add(new RomanMap(1,"I"));
        
        StringBuilder result = new StringBuilder();
        int lastPosition = 0; 
        
        while(num > 0 ){
            for(int i= lastPosition ;i < list.size();i++){
                RomanMap map = list.get(i);
                if(num >= map.getNumber()){
                    num -= map.getNumber();
                    result.append(map.getRomanNumber());
                    i--;
                    lastPosition = i;
                }
            }
        }
        return result.toString();
    }
}