Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
605 views
in Technique[技术] by (71.8m points)

language agnostic - How would you calculate all possible permutations of 0 through N iteratively?

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.

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…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

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

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…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...