g（n）复杂度为O（n ^ 2）的算法的时间复杂度

Question

I have just learnt about the time complexity of various sorting method. (eg Merge Sort, quick sort) However i am still a beginner at this field. I know that if g(n) has a complexity of O(n) the whole time complexity of this method would be n logn. But what if the complexity of g(n) is O(n^2)?

void f(n) { 
    if (n <= 1) return;
    else { 
      g(n); 
      f(n/2);
      f(n/2);
    }
}

Answer 1

The time complexity recurrence relation is

– where G(n) is the time complexity of g(n) . Methods to solve this for eg O(n^2) :

1. Expansion (dropping the O(...) until the end):

   

   – after m expansions. The second term contains a geometric series which converges to 2 , so it can be ignored as a constant. Applying the stopping condition n = 1 :

   

2. The Master Theorem . For the example:

   

   – which means Case 3 applies. Therefore the result is consistent with (1):