13. Roman to Integer

2024年1月20日 53次阅读

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

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

說一句廢話：本來這題不太想寫的，確實對羅馬數字不太熟悉，寫起來比較困難，不過最後想想還是寫吧，對於一個強迫症患者LeetCode已完成題目中間空一塊太難受了。

題目其實挺簡單的，無非就是將字符串分解，然後根據羅馬數字的規則，進行數字的轉換。

注意下面幾個規則就好了：

  • I (1)can be placed before V (5) and X (10) to make 4 and 9. 
  • X (5)can be placed before L (50) and C (100) to make 40 and 90. 
  • C (100)can be placed before D (500) and M (1000) to make 400 and 900.

注意到，如果羅馬數字左邊的數比右邊那個數要小，那麼需要將兩位數組合起來，右邊減掉左邊，爲轉換後的數。

fun romanToInt(s: String): Int {
    val sb = StringBuilder(s)
    var result = 0
    while (sb.isNotEmpty()){
        if (sb.length > 1 && romanToNumber(sb[0]) < romanToNumber(sb[1])){
            result += (romanToNumber(sb[1]) - romanToNumber(sb[0]))
            sb.delete(0, 2)
        }else{
            result += romanToNumber(sb[0])
            sb.deleteCharAt(0)
        }
    }
    return result
}

fun romanToNumber(c: Char): Int {
    return when (c) {
        'I' -> 1
        'V' -> 5
        'X' -> 10
        'L' -> 50
        'C' -> 100
        'D' -> 500
        'M' -> 1000
        else -> 0
    }
}

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