Question

I need to calculate permutations iteratively. The method signature looks like:

int[][] permute(int n)

For n = 3 for example, the return value would be:

[[0,1,2],
 [0,2,1],
 [1,0,2],
 [1,2,0],
 [2,0,1],
 [2,1,0]]

How would you go about doing this iteratively in the most efficient way possible? I can do this recursively, but I’m interested in seeing lots of alternate ways to doing it iteratively.

Answer 1

see QuickPerm algorithm, it’s iterative : http://www.quickperm.org/

Edit:

Rewritten in Ruby for clarity:

def permute_map(n)
  results = []
  a, p = (0...n).to_a, [0] * n
  i, j = 0, 0
  i = 1
  results << yield(a)
  while i < n
    if p[i] < i
      j = i % 2 * p[i] # If i is odd, then j = p[i], else j = 0
      a[j], a[i] = a[i], a[j] # Swap
      results << yield(a)
      p[i] += 1
      i = 1
    else
      p[i] = 0
      i += 1
    end
  end
  return results
end