[LeetCode] 78. Subsets tag: backtracking-优快云博客

本文详细介绍了如何通过深度优先搜索（DFS）的回溯方法来生成一个整数集合的所有可能子集（幂集）。该算法确保了结果中不包含重复的子集，并且考虑到了特殊情况。文中提供了两种实现方式：一种使用了深拷贝，另一种则避免了深拷贝的使用。

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],
  []
]

这个题目思路就是类似于DFS的backtracking.

1. Constraints

1) No duplicate subsets and neither integers in the nums.

2) edge case len(nums) == 0 => [ [ ] ] 但是还是可以

2. Ideas

DFS 的backtracking T: O(2^n) S; O(n)

3. Code

1) using deepcopy, but the idea is obvious, use DFS [] -> [1] -> [1, 2] -> [1, 2, 3] then pop the last one, then keep trying until the end.

import copy
class Solution:
    def subsets(self, nums: 'List [int]') -> 'List [ List [int] ]' :
        ans = []
        def helper(nums, ans, temp, pos):
            ans.append(copy.deepcopy(temp))
            for i in range(pos, len(nums)):
                temp.append()
                helper(nums, ans, temp, i + 1)
                temp.pop()
        
        helper(nums, ans, [], 0)
        return ans

2) Use the same idea , but get rid of deepcopy

class Solution:
    def subsets(self, nums: 'List[int]') -> 'List[List[int]]':
        ans = []
        def helper(nums, ans, temp, pos):
            ans.append(temp)
            for i in range(pos, len(nums)):
                helper(nums, ans, temp + [nums[i]], i + 1)
        helper(nums, ans, [], 0)
        return ans