LeetCode Integer to Roman（012）解法總結

描述

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

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"複製程式碼

Example 2:

Input: 4
Output: "IV"複製程式碼

Example 3:

Input: 9
Output: "IX"複製程式碼

Example 4:

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

Example 5:

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

思路

和vivo的數位之積一樣，是先從大的內容開始分解。不同之處在於，這個是減法，而數位之積是除法。

出去題目中給出的，四個連續的相同字元也要轉換為減法表達的形式。要在陣列加上對應的項，再直接遍歷即可。

class Solution {
    public String intToRoman(int num) {
    
    //設定對應的參考數值
    int[] value = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
    String[] symbol = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
        StringBuffer temp = new StringBuffer();
        int i = 0;

        //當分解後的數字不為0時，繼續分解。
        while(num != 0){
            if(num >= value[i]){
                temp.append(symbol[i]);
                num = num - value[i];
            }else{
                //當前值無法作為成分時，使用下一個值
                i++;
            }
        }
        return temp.toString();
    }
}複製程式碼

本題的難點在於找到完整的參考值序列。