47. Permutations II (Medium)

https://leetcode.com/problems/permutations-ii/

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

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

Solutions

1.

class Solution {
    Set<String> uniqueHash = new HashSet<>();

    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return ans;
        }

        Arrays.sort(nums);

        doPermutation(nums, ans, 0);

        return ans;
    }

    private void doPermutation(int[] nums, List<List<Integer>> ans, int depth) {
        int len = nums.length;

        if (depth == len - 1) {
            List<Integer> solution = new ArrayList<>();

            for (int n : nums) {
                solution.add(n);
            }

            if (uniqueHash.add(solution.toString())) {
                ans.add(solution);
            }

            return;
        }

        for (int j = depth; j < len; j++) {
            // pruning
            if (j != depth && nums[j] == nums[depth]) {
                continue;
            }

            swap(nums, depth, j);

            doPermutation(nums, ans, depth + 1);

            swap(nums, depth, j);
        }
    }

    void swap(int[] nums, int x, int y) {
        int tmp = nums[x];
        nums[x] = nums[y];
        nums[y] = tmp;
    }
}

2.

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();

        Arrays.sort(nums);

        doPermutation(nums, ans, 0);

        return ans;
    }

    void doPermutation(int[] nums, List<List<Integer>> ans, int depth) {
        if (depth == nums.length - 1) {
            List<Integer> list = new ArrayList<>();

            for (int n : nums) {
                list.add(n);
            }

            ans.add(list);

            return;
        }

        // From the perspective how you do permutation, first slot has len situations.
        // But the problem also points out the duplicate element issue, so the number of situations
        // is the count of unique element.
        // For elements already swapped, ignore it. Here we use a set to track the swapped value.

        Set<Integer> swaped = new HashSet<>();
        for (int j = depth, l = nums.length; j < l; j++) {
            if (!swaped.add(nums[j])) {
                continue;
            }

            // Above commented out code is not correct, should not be executed
            /*
            if (depth != 0 && nums[depth] == nums[depth-1]) {
                continue;
            }

            if (j != depth && nums[j] == nums[depth]) {
                continue;
            }

            if (j != depth && nums[j] == nums[j-1]) {
                continue;
            }
            */

            swap(nums, depth, j);

            doPermutation(Arrays.copyOf(nums, nums.length), ans, depth + 1);

            swap(nums, depth, j);
        }
    }

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

Incorrect Solutions

References