今天是紀錄LeetCode解題的第十二天

第十二題題目：Given an integer, convert it to a Roman numeral.
Roman numerals are formed by appending the conversions of decimal place values from highest to lowest. Converting a decimal place value into a Roman numeral has the following rules:

• If the value does not start with 4 or 9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.
• If the value starts with 4 or 9 use the subtractive form representing one symbol subtracted from the following symbol, for example, 4 is 1 (I) less than 5 (V): IV and 9 is 1 (I) less than 10 (X): IX. Only the following subtractive forms are used: 4 (IV), 9 (IX), 40 (XL), 90 (XC), 400 (CD) and 900 (CM).
• Only powers of 10 (I, X, C, M) can be appended consecutively at most 3 times to represent multiples of 10. You cannot append 5 (V), 50 (L), or 500 (D) multiple times. If you need to append a symbol 4 times use the subtractive form.

給定一個整數，將其轉換為羅馬數字
羅馬數字由小數位數值的轉換結果由高到低依序相加而成，將小數位數值轉換為羅馬數字有以下規則：

• 如果值不是以 4 或 9 開頭，則選擇可以從輸入中減去的最大值的符號，將該符號附加到結果，減去其值，然後將餘數轉換為羅馬數字。
• 如果數值以 4 或 9 開頭，則使用減法形式 ，表示從後面的符號中減去一個符號，例如，4比5（V）小1（I）：IV 9比 10（X）小 1（I）：IX，僅使用以下減法形式：4（IV）、9（IX）、40（XL）、90（XC）、400（CD）和 900（CM）。
• 只有 10 的冪次方（I、X、C、M）可以連續添加最多 3 次，以表示 10 的倍數。您不能多次新增 5 ( V)、50 ( L) 或 500 ( D)，如果需要增加 4 次符號，請使用減法形式

程式碼

class Solution:
    def intToRoman(self, num: int) -> str:
        val = [1000, 900, 500, 400,
               100, 90, 50, 40,
               10, 9, 5, 4,
               1]
        syms = ["M", "CM", "D", "CD",
                "C", "XC", "L", "XL",
                "X", "IX", "V", "IV",
                "I"]

        result = []
        for i, v in enumerate(val):
            while num >= v:
                num -= v
                result.append(syms[i])
        return "".join(result)

邏輯很簡單，就是當num大於該數字，就減掉，並把對應的羅馬符號加進result裡

執行過程

初始狀態

• num = 58

當i = 6、v = 50時，num > v

• num = num - v = 58 - 50 = 8
• result.append(syms[6]) → result = [" L "]

當i = 10、v = 5時，num > v

• num = num - v = 8 - 5 = 3
• result.append(syms[10]) → result = [" L ", " V "]

當i = 12、v = 1時，num > v

• num = num - v = 3 - 1 = 2
• result.append(syms[12]) → result = [" L ", " V ", " I "]
• 以上重複三次

最後就會得到result = [" L ", " V ", " I ", " I ", " I "]在join印出就是答案了