Reverse an ipairs table in Lua (not clever O(n) could be O(n/2) with swapping)

reverseTable.lua

--
-- Reverses an ipairs table
--
--local r = ReverseTable({'a', 'b', 'c'})
--for k, v in ipairs(r) do
--    print(k, v)
--end
--
--
function ReverseTable(t)
    local reversedTable = {}
    local itemCount = #t
    for k, v in ipairs(t) do
        reversedTable[itemCount + 1 - k] = v
    end
    return reversedTable
end

tohle funguje

^ agreed

If you don't mind modifying the original talbe, this should do the trick for O(n/2):

function reverse(tbl)
  for i=1, math.floor(#tbl / 2) do
    local tmp = tbl[i]
    tbl[i] = tbl[#tbl - i + 1]
    tbl[#tbl - i + 1] = tmp
  end
end

It's even a line shorter 😄

function reverse(tbl)
  for i=1, math.floor(#tbl / 2) do
    tbl[i], tbl[#tbl - i + 1] = tbl[#tbl - i + 1], tbl[i]
  end
end

Isn't this one working too?

O(n/2) is O(n)
I think you mean n/2, without the O

function table.reverse(t)
	local len = #t
	for i = len - 1, 1, -1 do
		t[len] = table.remove(t, i)
	end
end

or modified ipairs-function

function rpairs(t)
	return function(t, i)
		i = i - 1
		if i ~= 0 then
			return i, t[i]
		end
	end, t, #t + 1
end