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],
  [2,1,1]
]

java

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null) {
            return result;
        }
        if (nums.length == 0) {
            result.add(new ArrayList<Integer>());
            return result;
        }
        boolean[] visit = new boolean[nums.length];
        Arrays.sort(nums);
        dfs(nums, visit, new ArrayList<Integer>(), result);
        return result;
    }
    private void dfs(int[] nums, 
                     boolean[] visit,
                     List<Integer> path, 
                     List<List<Integer>> result) {
        if (path.size() == nums.length) {
            result.add(new ArrayList<Integer>(path));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (visit[i] == true) {
                continue;
            }
            if (i != 0 && nums[i] == nums[i - 1] && visit[i - 1] == false) {
                continue;
            }
            visit[i] = true;
            path.add(nums[i]);
            dfs(nums, visit, path, result);
            path.remove(path.size() - 1);
            visit[i] = false;
        }
    }
}

python

class Solution(object):
    def permuteUnique(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if nums is None:
            return []
        if len(nums) == 0:
            return [[]]
        visit, result = [False] * len(nums), []
        nums.sort()
        self.dfs(nums, visit, [], result)
        return result
        
    def dfs(self, nums, visit, path, result):
        if (len(path) == len(nums)):
            result.append(path[:])
            return
        for i in range(len(nums)):
            if visit[i] == True:
                continue
            if i != 0 and nums[i] == nums[i - 1] and visit[i - 1] == False:
                continue
            visit[i] = True
            path.append(nums[i])
            self.dfs(nums, visit, path, result)
            path.pop()
            visit[i] = False