[Swift]LeetCode587. 安装栅栏 | Erect the Fence
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➤微信公众号：山青咏芝（shanqingyongzhi）
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There are some trees, where each tree is represented by (x,y) coordinate in a two-dimensional garden. Your job is to fence the entire garden using the minimum length of rope as it is expensive. The garden is well fenced only if all the trees are enclosed. Your task is to help find the coordinates of trees which are exactly located on the fence perimeter. git
Example 1:github
Input: [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]] Output: [[1,1],[2,0],[4,2],[3,3],[2,4]] Explanation:
Example 2:微信
Input: [[1,2],[2,2],[4,2]] Output: [[1,2],[2,2],[4,2]] Explanation:
Even you only have trees in a line, you need to use rope to enclose them.
Note:app
- All trees should be enclosed together. You cannot cut the rope to enclose trees that will separate them in more than one group.
- All input integers will range from 0 to 100.
- The garden has at least one tree.
- All coordinates are distinct.
- Input points have NO order. No order required for output.
在一个二维的花园中，有一些用 (x, y) 坐标表示的树。因为安装费用十分昂贵，你的任务是先用最短的绳子围起全部的树。只有当全部的树都被绳子包围时，花园才能围好栅栏。你须要找到正好位于栅栏边界上的树的坐标。ui
示例 1:spa
输入: [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]] 输出: [[1,1],[2,0],[4,2],[3,3],[2,4]] 解释:
示例 2:3d
输入: [[1,2],[2,2],[4,2]] 输出: [[1,2],[2,2],[4,2]] 解释:
即便树都在一条直线上，你也须要先用绳子包围它们。
注意:code
- 全部的树应当被围在一块儿。你不能剪断绳子来包围树或者把树分红一组以上。
- 输入的整数在 0 到 100 之间。
- 花园至少有一棵树。
- 全部树的坐标都是不一样的。
- 输入的点没有顺序。输出顺序也没有要求。
1 /** 2 * Definition for a point. 3 * public class Point { 4 * public var x: Int 5 * public var y: Int 6 * public init(_ x: Int, _ y: Int) { 7 * self.x = x 8 * self.y = y 9 * } 10 * } 11 */ 12 class Solution { 13 func outerTrees(_ points: [Point]) -> [Point] { 14 var res:[Point] = [Point]() 15 var first:Point = points[0] 16 var firstIdx:Int = 0 17 var n:Int = points.count 18 for i in 1..<n 19 { 20 if points[i].x < first.x 21 { 22 first = points[i] 23 firstIdx = i 24 } 25 } 26 res.append(first) 27 var cur:Point = first 28 var curIdx:Int = firstIdx 29 while(true) 30 { 31 var next:Point = points[0] 32 var nextIdx:Int = 0 33 for i in 1..<n 34 { 35 if i == curIdx {continue} 36 var cross:Int = crossProduct(cur, points[i], next) 37 if nextIdx == curIdx || cross > 0 || (cross == 0 && dist(points[i], cur) > dist(next, cur)) 38 { 39 next = points[i] 40 nextIdx = i 41 } 42 } 43 for i in 0..<n 44 { 45 if i == curIdx {continue} 46 var cross:Int = crossProduct(cur, points[i], next) 47 if cross == 0 48 { 49 if check(&res, points[i]) 50 { 51 res.append(points[i]) 52 } 53 } 54 } 55 cur = next 56 curIdx = nextIdx 57 if curIdx == firstIdx {break} 58 } 59 return res 60 } 61 62 func crossProduct(_ A:Point,_ B:Point,_ C:Point) -> Int 63 { 64 var BAx:Int = A.x - B.x; 65 var BAy:Int = A.y - B.y; 66 var BCx:Int = C.x - B.x; 67 var BCy:Int = C.y - B.y; 68 return BAx * BCy - BAy * BCx; 69 } 70 71 func dist(_ A:Point,_ B:Point) -> Int 72 { 73 return (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y) 74 } 75 76 func check(_ res:inout [Point],_ p:Point) -> Bool 77 { 78 for r in res 79 { 80 if r.x == p.x && r.y == p.y {return false} 81 } 82 return true 83 } 84 }