Pattern: Permutations

A Permutations is rearrangement of all the elements of the array.
Example:
Permutations

class Solution {
    List<List<Integer>> list = new ArrayList<>();
    public List<List<Integer>> permute(int[] nums) {
        int taken[] = new int[nums.length];
        permute(nums, taken, new ArrayList<>());
        return list;
    }
    public void permute(int nums[], int [] taken, List<Integer> l){
        if(l.size()== nums.length){
            list.add(new ArrayList<>(l));
            return;
        }
        for(int i = 0;i< nums.length;i++){
            if(i>0 && nums[i]==nums[i-1] && taken[i-1]==1) continue;
            if(taken[i]==0){
                taken[i] = 1;
                l.add(nums[i]);
                permute(nums,taken,l);
                l.remove(l.size()-1);
                taken[i]=0;
            }
        }
    }
}

Example 2: Permutation II - get unique permutations

class Solution {
    List<List<Integer>> list = new ArrayList<>();
    public List<List<Integer>> permuteUnique(int[] nums) {
        Arrays.sort(nums);
        int taken[] = new int[nums.length];
        permute(nums, taken, new ArrayList<>());
        return list;  
    }
    public void permute(int nums[], int [] taken, List<Integer> l){
        if(l.size()== nums.length){
            list.add(new ArrayList<>(l));
            return;
        }
        for(int i = 0;i< nums.length;i++){
            //If previous duplicate is NOT used → skip current
            if(i>0 && nums[i]==nums[i-1] && taken[i-1]==0) continue;
            if(taken[i]==0){
                taken[i] = 1;
                l.add(nums[i]);
                permute(nums,taken,l);
                l.remove(l.size()-1);
                taken[i]=0;
            }
        }
    }
}