Leetcode 78 - 生成幂集时间复杂度

Question

Leetcode 78 question 可能的解决方案：

class Solution(object):
    def __init__(self):
        self.map = {}
    
    def helper(self, count, nums, vals, result):       
        
        if count == 0:
            result += [vals]        
            
        for i in range(len(nums)):
            self.helper(count - 1, nums[i+1:], vals + [nums[i]], result)
        
    
    def subsets(self, nums):
        result = []
        result.append([])
        
        for count in range(1,len(nums)+1):
            self.helper(count, nums, [], result)   
            
        return result

对于上面的解法，时间复杂度是O(2^n)还是O(n * 2^n)？

Answer 1

可以通过查找不同的 N 来找出复杂度 helper 函数被调用了多少次，代码方面如下所示：

class Solution(object):

    def __init__(self):
        self.map = {}
        self.counter = 0

    def helper(self, count, nums, vals, result):
        self.counter += 1
        if count == 0:
            result += [vals]

        for i in range(len(nums)):
            self.helper(count - 1, nums[i + 1:], vals + [nums[i]], result)

    def subsets(self, nums):
        result = [[]]

        for count in range(1, len(nums) + 1):
            self.helper(count, nums, [], result)

        return self.counter

因此：

N	Time the helper function gets called
1	2
2	8
3	24
4	64
5	160
6	384
7	896
8	2048
9	4608
10	10240
...	....
N	`O(n * 2^n)`

helper 函数的复杂度为 O(2^n)，因为您调用了列表的每个元素 nums:

for count in range(1,len(nums)+1):
    self.helper(count, nums, [], result)

整体时间复杂度为O(n * 2^n)

Leetcode 78 - 生成幂集时间复杂度

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