难度: Medium
原题连接
内容描述
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],
[]
]
思路 1
每次拿一个,跟res里面的每一个已有列表取并集再次插入res中
class Solution(object):
def subsets(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
res = [[]]
for num in nums:
res.extend([tmp+[num] for tmp in res])
return res
思路 2
BackTrack 标准解法版
对每个元素,有两种可能,加入 cur_lst 和不加入 cur_lst,写起来思路还是很清爽的
class Solution(object):
def subsets(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums.sort()
res = []
def search(cur_lst, idx):
if idx == len(nums):
res.append(cur_lst)
return
search(cur_lst + [nums[idx]], idx + 1)
search(cur_lst, idx + 1)
search([], 0)
return res
思路 3
DFS
class Solution(object):
def subsets(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums.sort()
res = []
def dfs(depth, start, lst):
res.append(lst)
if depth == len(nums):
return
for i in range(start, len(nums)):
dfs(depth+1, i+1, lst+[nums[i]])
dfs(0, 0, [])
return res