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 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
```