[Swift]LeetCode286. 牆和門 $ Walls and Gates

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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You are given a m x n 2D grid initialized with these three possible values.

1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

For example, given the 2D grid:

INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
  0  -1 INF INF

After running your function, the 2D grid should be:

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

---

您將獲得一個由這三個可能值初始化的m x n 2d網格。

1. -1 - 牆或障礙物。
2. 0 - 門.
3. INF - 無限意味着一個空房間。咱們使用值2^31-1=2147483647表示inf，由於您能夠假定到門的距離小於2147483647。

   把每一個空房間填滿到最近的門的距離。若是不可能到達一個門，應該用inf填充。

把每一個空房間填滿到最近的門的距離。若是不可能到達一個門，應該用inf填充。

例如，給定二維網格：

INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
  0  -1 INF INF

運行函數後，二維網格應爲：

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

---

Solution:BFS (Time Limit Exceeded)

 1 class Solution {
 2     func wallsAndGates(_ rooms:inout [[Int]]) {
 3         var q:[(Int,Int)] = [(Int,Int)]()
 4         var dirs:[[Int]] = [[0, -1],[-1, 0],[0, 1],[1, 0]]
 5         for i in 0..<rooms.count
 6         {
 7             for j in 0..<rooms[i].count
 8             {
 9                 if rooms[i][j] == 0
10                 {
11                     q.append((i, j))
12                 }
13             }
14         }
15         while(!q.isEmpty)
16         {
17             let i:Int = q.first!.0
18             let j:Int = q.first!.1
19             q.popLast()
20             for k in 0..<dirs.count
21             {
22                 let x:Int = i + dirs[k][0]
23                 let y:Int = j + dirs[k][1]
24                 if x < 0 || x >= rooms.count || y < 0 || y >= rooms[0].count || rooms[x][y] < rooms[i][j] + 1
25                 {
26                     continue
27                 }
28                 rooms[x][y] = rooms[i][j] + 1
29                 q.append((x, y))
30             }
31         }
32     }
33 }

---

Solution:DFS (嚴重超時產生錯誤)

 1 class Solution {
 2     func wallsAndGates(_ rooms:inout [[Int]]) {
 3         for i in 0..<rooms.count
 4         {
 5             for j in 0..<rooms[i].count
 6             {
 7                 if rooms[i][j] == 0
 8                 {
 9                     dfs(&rooms, i, j, 0)
10                 }
11             }
12         }
13     }
14     
15     func dfs(_ rooms:inout [[Int]],_ i:Int,_ j:Int,_ val:Int)
16     {
17         if i < 0 || i >= rooms.count || j < 0 || j >= rooms[i].count || rooms[i][j] < val
18         {
19             return
20         }
21         dfs(&rooms, i + 1, j, val + 1)
22         dfs(&rooms, i - 1, j, val + 1)
23         dfs(&rooms, i, j + 1, val + 1)
24         dfs(&rooms, i, j - 1, val + 1)
25     }
26 }

---

點擊：Playground測試

1 var sol = Solution()
2 let num:Int = Int(Int32.max - 1)
3 var arr:[[Int]] = [[num,-1,0,num],[num,num,num,-1],[num,-1,num,-1],[0,-1,num,num]]
4 sol.wallsAndGates(&arr)
5 print(arr)
6 //Print