QuickHull算法的复杂性？

Question

I know the complexity is O(nlog(n)). But why? How do you come to this answer?

Any help would be much appreciated, I'm very interested to know!

Answer 1

Its average case complexity is considered to be O(n log(n)) , whereas in the worst case it takes O(n^2) (quadratic).

Consider the following pseudo-code:

QuickHull (S, l, r)

     if S={ }    then return ()
else if S={l, r} then return (l, r)  // a single convex hull edge
else
    z = index of a point that is furthest (max distance) from xy.
    Let A be the set containing points strictly right of (x, z)
    Let B be the set containing points strictly right of (z, y)
    return {QuickHull (A, x, z) U (z) U QuickHull (B, z, y)}

The partition is determined by the line passing through two distinct extreme points: the rightmost lowest r and the leftmost highest points l . Finding the extremes require O(n) time.

For the recursive function, it takes n steps to determine the extreme point z , but the cost of recursive calls depends on the sizes of set A and set B .

Best case. Consider the best possible case, when each partition is almost balanced. Then we have

T(n) = 2 T(n/2) + O(n) .

This is a familiar recurrence relation, whose solution is

T(n) = O(n log(n)) .

This would occur with randomly distributed points.

Worst case. The worst case occurs when each partition is an extremely unbalanced. In that case the recurrence relation is

T(n) = T(n-1) + O(n) 
     = T(n-1) + cn

Repeated expansion shows this is O(n^2) . Therefore, in the worst case the QuickHull is quadratic.

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http://www.personal.kent.edu/~rmuhamma/Compgeometry/MyCG/ConvexHull/quickHull.htm