[Swift]LeetCode587. 安裝柵欄 | Erect the Fence

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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. 

Example 1:

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:

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:

1. All trees should be enclosed together. You cannot cut the rope to enclose trees that will separate them in more than one group.
2. All input integers will range from 0 to 100.
3. The garden has at least one tree.
4. All coordinates are distinct.
5. Input points have NO order. No order required for output.

---

在一個二維的花園中，有一些用 (x, y) 座標表示的樹。因爲安裝費用十分昂貴，你的任務是先用最短的繩子圍起全部的樹。只有當全部的樹都被繩子包圍時，花園才能圍好柵欄。你須要找到正好位於柵欄邊界上的樹的座標。

示例 1:

輸入: [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]]
輸出: [[1,1],[2,0],[4,2],[3,3],[2,4]]
解釋:

示例 2:

輸入: [[1,2],[2,2],[4,2]]
輸出: [[1,2],[2,2],[4,2]]
解釋:

即便樹都在一條直線上，你也須要先用繩子包圍它們。

注意:

1. 全部的樹應當被圍在一塊兒。你不能剪斷繩子來包圍樹或者把樹分紅一組以上。
2. 輸入的整數在 0 到 100 之間。
3. 花園至少有一棵樹。
4. 全部樹的座標都是不一樣的。
5. 輸入的點沒有順序。輸出順序也沒有要求。

---

Runtime: 292 ms

Memory Usage: 20.1 MB

 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 }