I need to calculate permutations iteratively. The method signature looks like:
int permute(int n)
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 QuickPerm algorithm, it's iterative : http://www.quickperm.org/
Rewritten in Ruby for clarity:
def permute_map(n) results =  a, p = (0...n).to_a,  * 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