Ruby所有可能的数组排列（一个内衬？）

Question

Questions similar to this have been asked before on SO, but they're not quite what I need and I can't seem to arrive at my solution through altering/modifying those approaches.

In any case, I have an array of arrays, as follows:

b= [["1"],["2"],["3"],["4"],["5"],["6"]]

(If it makes it easier to arrive at a solution, b can also be a one dimensional array, as follows: ["1","2","3","4","5","6"] . Either type of input works for my needs.)

and I would like to generate the following:

[["123456"],["213456"],["312456"],...] 

where each array in the output array is a unique permutation of the six numbers. I would also take it as a single array (eg, ["123456", "213456",...] ). The order of the output isn't particularly important as long as each entry is unique and no number repeats in a string (eg, "112345" isn't allowed). All 6 numbers must also be used in each entry, so I'm not interested in incremental output like "123" , either.

As much as this sounds like it, this isn't a homework problem. I could brute for this thing and get the output I need. I just feel like there has to be a better, more elegant, solution.

Answer 1

With Array#permutation :

permutations = (1..6).to_a.permutation.map(&:join)
# ["123456", "123465", "123546", ..., "654312", "654321"]

Answer 2

Ruby does this natively :) From the ruby documentation :

a = [1, 2, 3]
a.permutation.to_a     #=> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
a.permutation(1).to_a  #=> [[1],[2],[3]]
a.permutation(2).to_a  #=> [[1,2],[1,3],[2,1],[2,3],[3,1],[3,2]]
a.permutation(3).to_a  #=> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
a.permutation(0).to_a  #=> [[]] # one permutation of length 0
a.permutation(4).to_a  #=> []   # no permutations of length 4

http://www.ruby-doc.org/core-1.9.3/Array.html#method-i-permutation

Answer 3

You should definitely have a look at Permutation Gem . Example from documentation

perm = Permutation.new(3)
# => #<Permutation:0x57dc94 @last=5, @rank=0, @size=3>
colors = [:r, :g, :b]
# => [:r, :g, :b]
perm.map { |p| p.project(colors) }
# => [[:r, :g, :b], [:r, :b, :g], [:g, :r, :b], [:g, :b, :r], [:b, :r, :g],
#  [:b, :g, :r]]

UPDATE

If you are using Ruby > 1.8.6, Array.permutation is built in.

Answer 4

这应该这样做：

b.permutation.to_a.collect! { |i| i = [i.flatten.join] }