递归算法的时间复杂度

Question

I have a grid with -sided field in it. Every field contains a link to it's surrounding fields. [ ]

I have an algorithm which is implemented in this field, (which can probably be optimized):

[java like pseudocode]

public ArrayList getAllFields(ArrayList list) {

  list.addToList(this);

  for each side {
    if ( ! list.contains(neighbour) && constantTimeConditionsAreMet()) {
      neighbour.getAllFields(list) //Recursive call
    }
  }

  return list;

}

I'm having trouble finding the time complexity.

ArrayList#contains(Object) runs in linear time

How do i find the time complexity? My approach is this:

T(n) = O(1) + T(n-1) +
c(nbOfFieldsInArray - n) [The time to check the ever filling ArrayList]

T(n) = O(1) + T(n-1) + c*nbOfFieldsInArray - cn

Does this give me T(n) = T(n-1) + O(n) ?

Answer 1

The comment you added to your code is not helpful. What does getContinent do?

In any case, since you're using a linear search ( ArrayList.contains ) for every potential addition to the list, then it looks like the complexity will be Omega(n^2).

Answer 2

You recurrence seems correct T(n) = T(n-1) + theta(1) .

If you draw the recursion tree you'll notice you have a single branch with the values theta(n-1), theta(n-2), ..., theta(2), theta(1) , if you add up all the levels you get the arithmetic series 1+2+3+...+n

S1 = 1+2+3+...+n

If you define

S2 = n+...+3+2+1

and then calculate S1+S2 you get

S1 + S2 = 2*S1 = (n+1) + (n+1) + ... + (n+1) = n(n+1)

therefore

2*S1 = n(n-1) => S1 = n(n-1)/2

which means T(n) = 1/2 theta(n(n-1)) = 1/2 theta(n^2) = theta(n^2)