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Given an integer array nums
that may contain duplicates, return all possible subsets (the power set).
The solution set must not contain duplicate subsets. Return the solution in any order.
Example 1:
Input: nums = [1,2,2] Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]
Example 2:
Input: nums = [0] Output: [[],[0]]
Constraints:
1 <= nums.length <= 10
-10 <= nums[i] <= 10
class Solution:
def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
def dfs(nums, i, res, path):
res.append(copy.deepcopy(path))
for j in range(i, len(nums)):
if j != i and nums[j] == nums[j - 1]:
continue
path.append(nums[j])
dfs(nums, j + 1, res, path)
path.pop()
res, path = [], []
nums.sort()
dfs(nums, 0, res, path)
return res
class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<Integer> path = new ArrayList<>();
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(nums);
dfs(nums, 0, res, path);
return res;
}
private void dfs(int[] nums, int i, List<List<Integer>> res, List<Integer> path) {
res.add(new ArrayList<>(path));
for (int j = i; j < nums.length; ++j) {
if (j != i && nums[j] == nums[j - 1]) {
continue;
}
path.add(nums[j]);
dfs(nums, i + 1, res, path);
path.remove(path.size() - 1);
}
}
}
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