计算此特定代码的时间复杂度

Question

我试图找出以下代码的时间复杂度。 lst_of_lsts包含m个列表，其长度最多为n。

我认为：所有内容都在O(m*n)中运行，最小值为O(m*n)，remove = O(m*n)。

总体O(m*n)*O(m*n) = O(m^2*n^2)

我是对的吗？

 def multi_merge_v1(lst_of_lsts): 
      all = [e for lst in lst_of_lsts for e in lst] 
      merged = [] 
      while all != []: 
          minimum = min(all) 
          merged += [minimum] 
          all.remove(minimum) 
      return merged 

Answer 1

def multi_merge_v1(lst_of_lsts): # M lists of N
  all = [e for lst in lst_of_lsts for e in lst]
  # this is messy. Enumerate all N*M list items --> O(nm)

  merged = [] # O(1)
  while all != []: # n*m items initially, decreases in size by 1 each iteration.
    minimum = min(all) # O(|all|).
    # |all| diminishes linearly from (nm) items, so this is O(nm).
    # can re-use work here. Should use a sorted data structure or heap

    merged += [minimum] # O(1) amortized due to table doubling
    all.remove(minimum) # O(|all|)
  return merged # O(1)

整体复杂度似乎为O（nm + nm * nm + 1）= O（n ^ 2m ^ 2 + 1）。让人惊讶。