快速解释一下，我最近开始使用 codewars 来进一步提高我的编程技能，我的第一个挑战是制作一个罗马数字解码器，我经历了很多版本，因为我对我所拥有的不满意，所以我问是否有更简单的处理罗马数字所具有的所有模式的方法，例如 I 是 1，但如果 I 紧邻另一个数字，它会将其带走，例如 V = 5 但 IV = 4。
这是我的代码：
function Roman_Numerals_Decoder (roman)
local Dict = {I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000}
local number = 0
local i = 1
while i < #roman + 1 do
local letter = roman:sub(i,i) -- Gets the current character in the string roman
if roman:sub(i,i) == "I" and roman:sub(i + 1,i + 1) ~= "I" and roman:sub(i + 1,i + 1) ~= "" then -- Checks for the I pattern when I exists and next isnt I
number = number + (Dict[roman:sub(i +1,i + 1)] - Dict[roman:sub(i,i)]) -- Taking one away from the next number
i = i + 2 -- Increase the counter
else
number = number + Dict[letter] -- Adds the numbers together if no pattern is found, currently checking only I
i = i + 1
end
end
return number
end
print(Roman_Numerals_Decoder("MXLIX")) -- 1049 = MXLIX , 2008 = MMVIII目前我正试图让 1049 (MXLIX) 工作，但我得到了 1069，显然我没有遵守规则，我觉得它更错误，那应该是因为通常如果它不纠正它的 1 或 2 个数字是错误的.
回答1
该算法略有不同：当前一个字符的权重小于下一个字符时，您需要考虑减法。
function Roman_Numerals_Decoder (roman)
local Dict = {I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000}
local num = 0
local i = 1
for i=1, #roman-1 do
local letter = roman:sub(i,i) -- Gets the current character in the string roman
local letter_p = roman:sub(i+1,i+1)
if (Dict[letter] < Dict[letter_p]) then
num = num - Dict[letter] -- Taking one away from the next number
print("-",Dict[letter],num)
else
num = num + Dict[letter] -- Adds the numbers together if no pattern is found, currently checking only I
print("+",Dict[letter],num)
end
end
num = num + Dict[roman:sub(-1)];
print("+",Dict[roman:sub(-1)], num)
return num
end
print(Roman_Numerals_Decoder("MXLIX")) -- 1049 = MXLIX , 2008 = MMVIII