Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

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class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        dfs(list, new ArrayList<>(), nums, 0);
        return list;
    }
    
    public void dfs(List<List<Integer>> list, List<Integer> arr, int[] nums, int start) {
        list.add(new ArrayList<>(arr));
        for (int i = start; i < nums.length; ++i) {
            if (i > start && nums[i] == nums[i-1]) continue;
            arr.add(nums[i]);
            dfs(list, arr, nums, i + 1);
            arr.remove(arr.size() - 1);
        }
    }
}
  • Arrays.sort是为了后面remove duplicates;
  • 因为后面dfs(..., i+1)仍然是i+1, 故i和i-1比较，所以if去重条件是那样