Permutations of Array

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

Thoughts:

You take first element of an array (k=0) and exchange it with any element (i) of the array. Then you recursively apply permutation on array starting with second element. This way you get all permutations starting with i-th element. The tricky part is that after recursive call you must swap i-th element with first element back, otherwise you could get repeated values at the first spot. By swapping it back we restore order of elements (basically you do backtracking).

Code:

 public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> lists = new ArrayList<List<Integer>>();
        p(nums,0,lists);
        return lists;
    }
    public void p(int[] nums, int k,List<List<Integer>> lists){
        if(k>=nums.length){
            List<Integer> current = new ArrayList<Integer>();
            for (int a : nums) {
                current.add(a);
            }
            lists.add(current);
        }
        else{
        for(int i=k;i<nums.length;i++){
            swap(nums, i, k);
            p(nums,k+1,lists);
            swap(nums,i,k); // BackTracking
        }
        }
        
    }
    
    public void swap (int[] arr, int m, int n){
        int temp = arr[m];
        arr[m]=arr[n];
        arr[n]=temp;
    }