Python Leetcode 78-发电机组时间复杂性

Leetcode 78潜在解决方案：

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

---

对于上述解决方案，时间复杂度是O（2^n）还是O（n*2^n）？

可以通过查找不同的

n

来确定复杂度

帮助程序调用了多少次，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
调用helper函数的时间
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）