找到这两组项目的所有可能组合吗?Lua

Question

我在Haskell和Python等不同的编程语言中看到了这个问题的类似答案，但它们都使用了Lua没有的内置功能，所以请不要将这个问题标记为重复。

假设我有两张桌子，如下所示：

table1 = {A,B,C}
table2 = {D,E,F}  

我想找到所有独特的方法来匹配两个表中的项目，答案应该是(用非正式的符号(：

AD,BE,CF
AD,BF,CE
AE,BD,CF
AE,BF,CD
AF,BD,CE
AF,BE,CD  

因此答案将存储在表中，表[1]将是{{A, D}, {B, E}, {C, F}}，依此类推。
表的长度可以是任何值，但两者的大小相同。

Answer 1

我们可以通过归纳法(不是最快的方法，但很容易编写/理解(获得所有的shuffle

local function deepcopy(orig)
local copy
if type(orig) == 'table' then
copy = {}
for orig_key, orig_value in next, orig, nil do
copy[deepcopy(orig_key)] = deepcopy(orig_value)
end
setmetatable(copy, deepcopy(getmetatable(orig)))
else
copy = orig
end
return copy
end
local function get_shuffles(N)
if N == 1 then
return {{1}}
end
local shuffles = get_shuffles(N-1)
local result = {}
for index = 1, #shuffles do
local shuffle = shuffles[index]
for position = 1, #shuffle do
local new_shuffle = deepcopy(shuffle)
table.insert(new_shuffle, position, N)
table.insert(result, new_shuffle)
end
local new_shuffle = deepcopy(shuffle)
table.insert(new_shuffle, N)
table.insert(result, new_shuffle)
end
return result
end
table1 = {"A", "B", "C"}
table2 = {"D","E", "F"}
assert(#table1 == #table2)

local result = {}
local shuffles = get_shuffles(#table1)
for index = 1, #shuffles do
local shuffle = shuffles[index]
local part = {}
for i = 1, 3 do
table.insert(part, {})
table.insert(part[i], table1[i])
table.insert(part[i], table2[shuffle[i]])
end
table.insert(result, part)
end
for index = 1, #result do
print(result[index][1][1], result[index][1][2], result[index][2][1], result[index][2][2], result[index][3][1], result[index][3][2])
end

Answer 2

function get_all_combinations(arr1, arr2)
local n, e, all_comb  = #arr1, {}, {}
for j = 1, n do
e[j] = arr2[j]
end
local function generate(m)
if m <= 1 then
local comb = {}
all_comb[#all_comb + 1] = comb
for j = 1, n do
comb[j] = arr1[j]..e[j]  -- it should be {arr1[j], e[j]} to fulfill your requirements
end
else
for j = 1, m do
generate(m - 1)
local k = j < m and m % 2 == 1 and 1 or j
e[k], e[m] = e[m], e[k]
end
end
end
generate(n)
return all_comb
end
for i, v in ipairs(get_all_combinations({"A", "B", "C"}, {"D", "E", "F"})) do
print(i, table.concat(v, ";"))
end

Answer 3

另一种方法是使用以下代码。这是为了帮助游戏(Typeshift(发现可变字母组的所有可能组合而写的。不过，我已经对它进行了修改，以适合你的例子。

-- table array: { {1, 2}, {3, 4}, {5, 6} }
-- Should return { 135, 136, 145, 146, 235, 236, 245, 246 }
--
-- This uses tail recursion so hopefully lua is smart enough not to blow the stack
function arrayCombine(tableArray)
-- Define the base cases
if (tableArray == nil) then
return nil
elseif (#tableArray == 0) then
return {}
elseif (#tableArray == 1) then
return tableArray[1]
elseif (#tableArray == 2) then
return arrayCombine2(tableArray[1], tableArray[2])
end -- if
-- We have more than 2 tables in the input parameter.  We want to pick off the *last*
-- two arrays, merge them, and then recursively call this function again so that we 
-- can work our way up to the front.
local lastArray = table.remove(tableArray, #tableArray)
local nextToLastArray = table.remove(tableArray, #tableArray)
local mergedArray = arrayCombine2(nextToLastArray, lastArray)
table.insert(tableArray, mergedArray)
return arrayCombine(tableArray)
end -- arrayCombine

function arrayCombine2(array1, array2)
local mergedArray = {}
for _, elementA in ipairs(array1) do
for _, elementB in ipairs(array2) do
table.insert(mergedArray, elementA .. elementB) 
end -- for
end -- for
return mergedArray
end -- arrayCombine2
-- You can set it up this way:
combinedArray = {}
table.insert(combinedArray, {"A", "B", "C"})
table.insert(combinedArray, {"D", "E", "F"})
for i,v in ipairs(arrayCombine(combinedArray)) do
print(i,v)
end
-- Or go this way, which may be somewhat cleaner:
for i,v in ipairs(arrayCombine({{"A", "B", "C"}, {"D", "E", "F"}})) do
print(i,v)
end

无论哪种方式，它都会产生你想要的结果。