# Will this solution has equal space complexity of the recursion one?

• Here is my iterative solution -- for i = 1 to n-1, each loop will generate all permutations of length i + 1 based on the permutations of length i. Will the space complexity be worse or be better than a recursive solution? Could it be improved somehow?

`````` public List<List<Integer>> permute(int[] num) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (num == null || num.length == 0) // return if trivial
return result;
// add num[0] into result list, since it's the only permutation case of length 1
List<Integer> list = new ArrayList<Integer>();
int n = num.length;
for (int i = 1; i < n; i++) // for v = num[1], num[2], ... num[n-1],
{
int v = num[i];
int m = result.size();
// generate all permutations of length i+1
for (int j = 0; j < m; j++)
{
list = result.get(j);
int o = list.size();
// by inserting v into each possible insertion point k of every permutations of length i
for(int k = 0; k <= o; k++)
{
ArrayList<Integer> permutation = new ArrayList<Integer>(list);