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Roman to Integer

#13.Roman to Integer

Problem statement

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
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For example, 2 is written as II in Roman numeral, just two ones 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 a roman numeral, convert it to an integer.

Example 1

Input: s = "III"
Output: 3
Explanation: III = 3.
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Example 2

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
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Example 3

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

給定一個羅馬數字將其轉換成阿拉伯數字,羅馬數字通常會從左到右、由大到小來撰寫,但有一特性,數字 4 不寫成 IIII 而是 IV,因為 45 之前,所以用 V 減去一個 I 得到 IV 也就是等於 4

相同原則下也是適用於 9,轉換成羅馬數字為 IX,題目也提供了六種使用這種減法的案例。

Solution

  • 先過濾特定條件,判斷字串 s 是否為空字串或是 null,如果成立則返回 0
  • 接著將羅馬符號對應的數字分別裝進 Dictionary 供後續轉換使用
  • 建立一個 counter 變數儲存加總,並先 assign s0 個索引轉換後的值
  • 迴圈走訪字串內的從所有字符(除了第一個字),用一個 If Statements 來判斷是加法還是減法,正常原則是由大到小,所以條件可設為這次的羅馬符號大於上次則為減法
    • 條件成立:加總這次轉換的數值
    • 條件不成立:邏輯是減去上次加總的,再加上這次的數值減去上次的數值
  • 最後返回結果 counter
public int RomanToInt(string s)
{
    if (s == null || s == String.Empty)
        return 0;

    IDictionary dic = new Dictionary<char, int>
    {
        { 'I', 1 },
        { 'V', 5 },
        { 'X', 10 },
        { 'L', 50 },
        { 'C', 100 },
        { 'D', 500 },
        { 'M', 1000 }
    };

    int counter = (int)dic[s[0]];

    for (int i = 0; i < s.Length; i++)
    {
        if ((int)dic[s[i - 1]] >= (int)dic[s[i]])
            counter += (int)dic[s[i]];
        else
            counter += (int)dic[s[i]] - (int)dic[s[i - 1]] * 2;
    }

    return counter;
}
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Reference

LeetCode Solution

GitHub Repository


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