LeetCode 12.Integer to Roman 阿拉伯數字轉換爲羅馬數字

LeetCode 12.Integer to Roman 阿拉伯數字轉換爲羅馬數字

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.

遞歸

通過和羅馬數字比較進行轉換。但是遞歸效率會很低，很多重複的計算。

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) : "";
    }
}

直接轉換

遞歸效率很低，我們可以採用迭代的方式根據轉換規則進行轉換。將每一個規則看成是一個鍵值對。建立一個RomanMap對象。然後根據給的數字的大小，與鍵值對的阿拉伯數字從大到小地進行對比，然後進行轉化。這裏設置了一個lastPosition的變量來減少比對的順序。雖然通過建立對象的方式效率會低一點，但是代碼的可讀性會增加。

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();
    }
}