LeetCode 47 - Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

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

 

Solution 1:

public List<List<Integer>> permuteUnique(int[] num) {
    List<List<Integer>> result = new ArrayList<>();
    Arrays.sort(num);
    permutate(result, num, 0);
    return result;
}

public void permutate(List<List<Integer>> result, int[] num, int pos) {
    if(pos == num.length) {
        List<Integer> list = new ArrayList<Integer>();
        for(int a:num) list.add(a);
        result.add(list);
        return;
    }
    for(int i=pos; i<num.length; i++) {
        if(hasDuplicate(num, pos, i)) continue;
        swap(num, i, pos);
        permutate(result, num, pos+1);
        swap(num, i, pos);
    }
}

private boolean hasDuplicate(int[] num, int start, int end) {
    for(int i=start; i<end; i++) {
        if(num[i] == num[end]) return true;
    }
    return false;
}

private void swap(int[] num, int i, int j) {
    int tmp = num[i];
    num[i] = num[j];
    num[j] = tmp;
}

 

Solution 2:

public List<List<Integer>> permuteUnique(int[] num) {
    List<List<Integer>> result = new ArrayList<>();
    result.add(new ArrayList<Integer>());
    Arrays.sort(num);
    for(int i=0; i<num.length; i++) {
        Set<List<Integer>> currentSet = new HashSet<List<Integer>>();
		for (List<Integer> l : result) {
			for (int j = 0; j < l.size() + 1; j++) {
				List<Integer> entry = new ArrayList<Integer>(l);
				entry.add(j, num[i]);
				currentSet.add(entry);
			}
		}
		result = new ArrayList<List<Integer>>(currentSet);
    }
    return result;
}