LeetCode_12_IntegerToRoman

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leetcode algorithm テキストリンク
LeetCode_12_IntegerToRoman
Problem
https://leetcode.com/problems/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, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII , which is simply X + II . The number 27 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.
    Example 1:
    Input: num = 3
Output: "III"
    Example 2:
    Input: num = 4
Output: "IV"
    Example 3:
    Input: num = 9
Output: "IX"
    Example 4:
    Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
    Example 5:
    Input: num = 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
    Constraints:
  • 1 <= num <= 3999
    • Solution
    もんだいぶんせき

  • アラビア数字からローマ数字への変換問題
  • IがVまたはXの前にある場合、4および9を作成することができる.
  • XがLまたはCの前にある場合、40および90を作成することができる.
  • CがDまたはMの前にある場合、400および900を作成することができる.

  • 各変換テーブルを以下に示します.
    Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

  • ただし、入力値numの範囲:1 <= num <= 3999
    • 方法.
    まず、1桁(1000桁を超えることはできません)に分け、各桁で構成部品を決定します.
    各ビット数にテーブルを作成し、[key:value]を使用して値を取得します.
    いずれにしても与えられた範囲は1 <= num <= 3999であり,代入は容易である.
  • で指定された値numを配列に変更します.ex) 1994 -> arr = [1, 9, 9, 4]
  • の各ビット数が表す数字はtableで見つかった.arr[0]は1000を表すので、「M」を検索します.
  • の結果に値を加算して返します.
    • class Solution {
    let table: [Int: String] = [
        0: "",
        1: "I",
        2: "II",
        3: "III",
        4: "IV",
        5: "V",
        6: "VI",
        7: "VII",
        8: "VIII",
        9: "IX",
        10: "X",
        20: "XX",
        30: "XXX",
        40: "XL",
        50: "L",
        60: "LX",
        70: "LXX",
        80: "LXXX",
        90: "XC",
        100: "C",
        200: "CC",
        300: "CCC",
        400: "CD",
        500: "D",
        600: "DC",
        700: "DCC",
        800: "DCCC",
        900: "CM",
        1000: "M",
        2000: "MM",
        3000: "MMM"
    ]
    
    private func digitCount(_ num: Int) -> Int{
        var n = num
        var cnt = 0
        while n >= 10{
            n = n / 10
            cnt += 1
        }
        
        return cnt
    }
    
    private func digits(_ num: Int) -> [Int]{
        var digits: [Int] = []
        var n = num
        digits.append(n % 10)
        while n >= 10 {
            n = n / 10
            digits.append(n % 10)
        }
        return digits.reversed()
    }
    
    private func pow(_ base: Int, _ exponent: Int) -> Int{
        var result = 1
        
        for _ in 0 ..< exponent{
            result = result * base
        }
        
        return result
    }
    
    func intToRoman(_ num: Int) -> String {
        var result: String = ""
        
        let arr = digits(num)
        for i in 0 ..< arr.count{
            let k = arr[i] * pow(10, arr.count-i-1)
            result += table[k]!
        }
        
        return result
    }
}

let solution = Solution()
print(solution.intToRoman(10))
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