Woodstock Blog

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[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].


Frequency 2
Difficulty 4
Adjusted Difficulty 4
Time to use --------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)


This question is based on “Permutations”, plus duplication avoidance.

This question, together with “Permutations”, is very classic and frequent questions, thus the basis for many similar DFS problems. Read this blog for more.

The idea is when getting element from remaining list and add to current list, check for duplication. If the number occured before, skip operation. (Don’t forget sorting is required at first).

The key points to keep in mind:

  1. Use another array to flag the ‘visited’ items
  2. Check items with same value, and ONLY USE THE FIRST INSTANCE OF THE SAME VALUE. Which is to say, if current = previous, but previous is not visited, do not use current number.

My code

public class Solution {
    public List<List<Integer>> permuteUnique(int[] num) {
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        if (num == null || num.length == 0) {
            return ans;
        int len = num.length;
        helper(ans, new ArrayList<Integer>(), num, len, new boolean[len]);
        return ans;

    private void helper(List<List<Integer>> ans, List<Integer> path, int[] num, int len, boolean[] visited) {
        if (path.size() == len) {
            ans.add(new ArrayList<Integer>(path));
        for (int i = 0; i < len; i++) {
            if (i > 0 && num[i - 1] == num[i] && !visited[i - 1]) {
                // important: if previous number have same value as (i)th
                // but have never been visited, then skip current number
            if (!visited[i]) {
                visited[i] = true;
                helper(ans, path, num, len, visited);
                path.remove(path.size() - 1);
                visited[i] = false;