587. Erect the Fence
利用一种算法叫做Monotone Chain，加上之前的旋转卡壳。。。还有一道求是否所有点形成convex hull 这三个解法好好搞一搞清楚
def outerTrees(self, points):
"""Computes the convex hull of a set of 2D points.
Input: an iterable sequence of (x, y) pairs representing the points.
Output: a list of vertices of the convex hull in counter-clockwise order,
starting from the vertex with the lexicographically smallest coordinates.
Implements Andrew's monotone chain algorithm. O(n log n) complexity.
"""
# Sort the points lexicographically (tuples are compared lexicographically).
# Remove duplicates to detect the case we have just one unique point.
# points = sorted(set(points))
points = sorted(points, key=lambda p: (p.x, p.y))
# Boring case: no points or a single point, possibly repeated multiple times.
if len(points) <= 1:
return points
# 2D cross product of OA and OB vectors, i.e. z-component of their 3D cross product.
# Returns a positive value, if OAB makes a counter-clockwise turn,
# negative for clockwise turn, and zero if the points are collinear.
def cross(o, a, b):
# return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])
return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x)
# Build lower hull
lower = []
for p in points:
while len(lower) >= 2 and cross(lower[-2], lower[-1], p) < 0:
lower.pop()
lower.append(p)
# Build upper hull
upper = []
for p in reversed(points):
while len(upper) >= 2 and cross(upper[-2], upper[-1], p) < 0:
upper.pop()
upper.append(p)
# Concatenation of the lower and upper hulls gives the convex hull.
# Last point of each list is omitted because it is repeated at the
# beginning of the other list.
# return lower[:-1] + upper[:-1]
return list(set(lower[:-1] + upper[:-1]))
