原题

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]
Output:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

解法1

使用组合, 从列表里分别取0,1,…n个数, 将结果放到res里, 由于itertools.combination返回的是tuple, 我们需要转化为list.
Time: O(n) + O(m), m是组合列表的长度
Space: O(1)

class Solution:
    def subsets(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        for i in range(len(nums)+1):
            res.extend(list(itertools.combinations(nums, i)))
        res = [list(tup) for tup in res]
        return res

解法2

深度优先搜索, 从空集开始, 依次将path放入res里

Time: O(n) + O(m), m是组合列表的长度
Space: O(1)

class Solution:
    def subsets(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """            
        res = []
        self.dfs(nums, 0, [], res)
        return res
    
    def dfs(self, nums, index, path, res):
        res.append(path)
        for i in range(index, len(nums)):
            self.dfs(nums, i+1, path+[nums[i]], res)