I have an array of ids
a1 = [1, 2, 3, 4, 5]
and I have another array of objects with ids in random order
a2 = [(obj_with_id_5), (obj_with_id_2), (obj_with_id_1), (obj_with_id_3), (obj_with_id_4)]
Now I need to sort a2 according to the order of ids in a1. So a2 should now become:
[(obj_with_id_1), (id_2), (id_3), (id_4), (id_5)]
a1 might be [3, 2, 5, 4, 1] or in any order but a2 should correspond to the order of ids in a1.
I do like this:
a1.each_with_index do |id, idx|
found_idx = a1.find_index { |c| c.id == id }
replace_elem = a2[found_idx]
a2[found_idx] = a2[idx]
a2[idx] = replace_elem
end
But this still might run into an O(n^2) time if order of elements of a2 is exactly reverse of a1. Can someone please tell me the most efficient way of sorting a2?
See Question&Answers more detail:
os 与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…