README_EN.md

90. Subsets II

中文文档

Description

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

Solutions

Python3

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

Java

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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