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.
Example:
Input: [1,2,2] Output: [ [2], [1], [1,2,2], [2,2], [1,2], [] ]
class Solution(object): def subsetsWithDup(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ res, lst = [], [] if nums is None or len(nums) == 0: return res nums.sort() self.dfs(nums, 0, lst, res) return res def dfs(self, nums, level, lst, res): if level == len(nums): res.append(list(lst)) return lst.append(nums[level]) self.dfs(nums, level + 1, lst, res) lst.pop() # after adding the char, skip the following same one while level < len(nums) - 1 and nums[level] == nums[level + 1]: level += 1 self.dfs(nums, level + 1, lst, res)
原文：https://www.cnblogs.com/xuanlu/p/11711339.html