您将如何迭代计算从0到N的所有可能排列? [英] How would you calculate all possible permutations of 0 through N iteratively?

问题描述

我需要迭代计算置换.方法签名如下:

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

int[][] permute(int n)

例如对于n = 3，返回值将是:

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]]

您将如何以最有效的方式进行迭代?我可以递归执行此操作，但是我有兴趣看到许多其他的迭代方法.

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.

推荐答案

请参阅QuickPerm算法，它是迭代的: http://www.quickperm.org/

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

为清楚起见，在Ruby中进行了重写:

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

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