Leetcode

Subsets II

  • Time:O(n \cdot 2^n)
  • Space:O(n \cdot 2^n)

C++

class Solution {
 public:
  vector<vector<int>> subsetsWithDup(vector<int>& nums) {
    vector<vector<int>> ans;
    sort(begin(nums), end(nums));
    dfs(nums, 0, {}, ans);
    return ans;
  }

 private:
  void dfs(const vector<int>& nums, int s, vector<int>&& path,
           vector<vector<int>>& ans) {
    ans.push_back(path);

    for (int i = s; i < nums.size(); ++i) {
      if (i > s && nums[i] == nums[i - 1])
        continue;
      path.push_back(nums[i]);
      dfs(nums, i + 1, move(path), ans);
      path.pop_back();
    }
  }
};

JAVA

class Solution {
  public List<List<Integer>> subsetsWithDup(int[] nums) {
    List<List<Integer>> ans = new ArrayList<>();
    Arrays.sort(nums);
    dfs(nums, 0, new ArrayList<>(), ans);
    return ans;
  }

  private void dfs(int[] nums, int s, List<Integer> path, List<List<Integer>> ans) {
    ans.add(new ArrayList<>(path));

    for (int i = s; i < nums.length; ++i) {
      if (i > s && nums[i] == nums[i - 1])
        continue;
      path.add(nums[i]);
      dfs(nums, i + 1, path, ans);
      path.remove(path.size() - 1);
    }
  }
}

Python

class Solution:
  def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
    ans = []

    def dfs(s: int, path: List[int]) -> None:
      ans.append(path)
      if s == len(nums):
        return

      for i in range(s, len(nums)):
        if i > s and nums[i] == nums[i - 1]:
          continue
        dfs(i + 1, path + [nums[i]])

    nums.sort()
    dfs(0, [])
    return ans