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GiftWrapping.kt
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GiftWrapping.kt
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/*
* Copyright (c) 2017 Kotlin Algorithm Club
*
* Permission is hereby granted, free of charge, to any person obtaining a copy
* of this software and associated documentation files (the "Software"), to deal
* in the Software without restriction, including without limitation the rights
* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
* copies of the Software, and to permit persons to whom the Software is
* furnished to do so, subject to the following conditions:
*
* The above copyright notice and this permission notice shall be included in all
* copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*/
package io.uuddlrlrba.ktalgs.geometry.convexhull
import io.uuddlrlrba.ktalgs.datastructures.Stack
import io.uuddlrlrba.ktalgs.geometry.Point
class GiftWrapping: ConvexHullAlgorithm {
override fun convexHull(points: Array<Point>): Collection<Point> {
if (points.size < 3) throw IllegalArgumentException("there must be at least 3 points")
val hull = Stack<Point>()
// Find the leftmost point
var l = 0
points.indices
.asSequence()
.filter { points[it].x < points[l].x }
.forEach { l = it }
// Start from leftmost point, keep moving counterclockwise
// until reach the start point again. This loop runs O(h)
// times where h is number of points in result or output.
var p = l
var q: Int
do {
// Add current point to result
hull.push(points[p])
// Search for a point 'q' such that orientation(p, x,
// q) is counterclockwise for all points 'x'. The idea
// is to keep track of last visited most counterclock-
// wise point in q. If any point 'i' is more counterclock-
// wise than q, then update q.
q = ( p+ 1) % points.size
points.indices
.asSequence()
.filter { Point.orientation(points[p], points[it], points[q]) < 0 }
.forEach { q = it }
// Now q is the most counterclockwise with respect to p
// Set p as q for next iteration, so that q is added to
// result 'hull'
p = q
} while (p != l) // While we don't come to first point
return hull
}
}