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469 Convex Polygon

469. Convex Polygon

题目: <

难度 : Medium



记得讲过convex hull,然而已经不知道是啥了。。。。


  • The polygon is entirely contained in a closed half-plane defined by each of its edges.

这个算法很有意思也很make sense,叫包礼物。不过也是对付convex hull的,看到有人提这个:

You can make things a lot easier than the Gift-Wrapping Algorithm... that's a good answer when you have a set of points w/o any particular boundary and need to find the convex hull.

A polygon is a set of points in a list where the consecutive points form the boundary. It is much easier to figure out whether a polygon is convex or not (and you don't have to calculate any angles, either):

For each consecutive pair of edges of the polygon (each triplet of points), compute the z-component of the cross product of the vectors defined by the edges pointing towards the points in increasing order. Take the cross product of these vectors:

The polygon is convex if the z-components of the cross products are either all positive or all negative. Otherwise the polygon is nonconvex.

given p[k], p[k+1], p[k+2] each with coordinates x, y:

dx1 = x[k+1]-x[k]

dy1 = y[k+1]-y[k]

dx2 = x[k+2]-x[k+1]

dy2 = y[k+2]-y[k+1]

zcrossproduct = dx1 * dy2 - dy1 * dx2

If there are N points, make sure you calculate N cross products, e.g. be sure to use the triplets (p[N-2],p[N-1],p[0]) and (p[N-1],p[0],p[1]).


class Solution(object):
    def isConvex(self, points):
        :type points: List[List[int]]
        :rtype: bool
        n = len(points)
        zcrossproduct = None

        for i in range(-2, n-2):
            x = [ points[i][0], points[i+1][0], points[i+2][0] ]
            y = [ points[i][1], points[i+1][1], points[i+2][1] ]

            dx1 = x[1] - x[0]
            dy1 = y[1] - y[0]

            dx2 = x[2] - x[1]
            dy2 = y[2] - y[1]

            if not zcrossproduct:
                zcrossproduct = dx1 * dy2 - dy1 * dx2
            elif ( dx1 * dy2 - dy1 * dx2 ) * zcrossproduct < 0:
                return False
        return True