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subsets.py
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44 lines (30 loc) · 944 Bytes
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'''
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],
[]
]
'''
class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
# Approach one 将nums拆解,递归求解, 相当于每次将nums[0] 添加到其他所有的已有子集中。
# if not nums: return [[]]
# res = self.subsets(nums[1:])
# return res + [[nums[0]] + s for s in res]
# Approach two 迭代求解, 消耗栈空间少
# res = [[]]
# for i in nums:
# res += [ n + [i] for n in res]
# return res
# Approach three 华丽但不好想到的位运算处理法, 效率较低
return [[nums[j] for j in range(len(nums)) if i>>j&1] for i in range(2**len(nums))]