If one only wants a convex hull, then one can use several algorithms, especially in 2D.
A simple one is Jarvis's march or gift wrapping. One starts with the farthest point from the center or the farthest point in some direction, then gives it a vector for what makes it the farthest. One then finds the angles of the other points to that vector, and selects the point with the lowest angle in some direction. One adds it to the hull, then uses the direction to it as its vector and repeats until one gets the first point again. Time: O(n*h) for n points and h hull points.
Graham's scan is like gift wrapping, but works with points sorted in order of their angles relative to the first point and its vector. One searches for each new hull point in the direction of increasing angle. Time: O(n*log

)
Quickhull is like quicksort. One starts with the two most separated points along one direction, connects them with a line, then finds the point that is the farthest from that line. All three points connected form a triangle, the initial hull. Drop the points inside the triangle from further consideration. For each edge, look for the farthest outside point, and add that point to the hull. The edge with that point forms a triangle, and drop all points inside of it. Repeat with each edge until no more points can be added to the hull. Time: O(n*log

) on average, O(n^2) in the worst case, like quicksort.
Divide and conquer. Like for Delaunay triangulation, except that one calculates a hull for each part. One then connects the two hulls to make the overall one.
Andrew's monotone chain, like Graham's scan, uses sorted vertices, but sorted along some linear direction, like a coordinate axis. One first finds the points with minimum and maximum position along that direction, then constructs a half-hull on each side of the line between the two points. Combining the two half-hulls gives the overall one.
The Kirkpatrick–Seidel algorithm, "the ultimate planar convex hull algorithm", is something like divide and conquer, but I have not been able to interpret it any further.
Chan's algorithm consists of dividing up the points into several subsets, finding the convex hull from each subset, and then finding the overall convex hull by treating these hulls' points as input points.
There is a nice trick for reducing the number of points that one needs to look at. In 2D, it's to remove the points inside of the quad (max x) - (max y) - (min x) - (min y). One can do similar things in more dimensions, like remove a similar sort of irregular octahedron in 3D.
Some of these algorithms can be generalized to more dimensions, like gift wrapping.