[Swift]LeetCode417. 太平洋大西洋水流問題 | Pacific Atlantic Water Flow

原文地址：http://www.javashuo.com/article/p-pghwkuuy-cc.html 

Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.

Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.

Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.

Note:

1. The order of returned grid coordinates does not matter.
2. Both m and n are less than 150. 

Example:

Given the following 5x5 matrix:

  Pacific ~   ~   ~   ~   ~ 
       ~  1   2   2   3  (5) *
       ~  3   2   3  (4) (4) *
       ~  2   4  (5)  3   1  *
       ~ (6) (7)  1   4   5  *
       ~ (5)  1   1   2   4  *
          *   *   *   *   * Atlantic

Return:

[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).

---

給定一個 m x n 的非負整數矩陣來表示一片大陸上各個單元格的高度。「太平洋」處於大陸的左邊界和上邊界，而「大西洋」處於大陸的右邊界和下邊界。

規定水流只能按照上、下、左、右四個方向流動，且只能從高到低或者在同等高度上流動。

請找出那些水流既能夠流動到「太平洋」，又能流動到「大西洋」的陸地單元的座標。 

提示：

1. 輸出座標的順序不重要
2. m 和 n 都小於150 

示例： 

給定下面的 5x5 矩陣:

  太平洋 ~   ~   ~   ~   ~ 
       ~  1   2   2   3  (5) *
       ~  3   2   3  (4) (4) *
       ~  2   4  (5)  3   1  *
       ~ (6) (7)  1   4   5  *
       ~ (5)  1   1   2   4  *
          *   *   *   *   * 大西洋

返回:

[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (上圖中帶括號的單元).

---

320ms

 1 class Solution {
 2     func pacificAtlantic(_ matrix: [[Int]]) -> [[Int]] {
 3         var matrix = matrix
 4         if matrix.isEmpty || matrix[0].isEmpty {return []}
 5         var res:[[Int]] = [[Int]]()
 6         var m:Int = matrix.count
 7         var n:Int = matrix[0].count
 8         var pacific:[[Bool]] = [[Bool]](repeating:[Bool](repeating:false,count:n),count:m)
 9         var atlantic:[[Bool]] = [[Bool]](repeating:[Bool](repeating:false,count:n),count:m)
10         for i in 0..<m
11         {
12             dfs(&matrix, &pacific, Int.min, i, 0)
13             dfs(&matrix, &atlantic, Int.min, i, n - 1)
14         }
15         for i in 0..<n
16         {
17             dfs(&matrix, &pacific, Int.min, 0, i)
18             dfs(&matrix, &atlantic, Int.min, m - 1, i)
19         }
20         
21         for i in 0..<m
22         {
23             for j in 0..<n
24             {
25                 if pacific[i][j] && atlantic[i][j]
26                 {
27                     res.append([i,j])
28                 }                
29             }           
30         }
31         return res
32     }
33     
34     func dfs(_ matrix:inout [[Int]],_ visited:inout [[Bool]],_ pre:Int,_ i:Int,_ j:Int)
35     {
36         var m:Int = matrix.count
37         var n:Int = matrix[0].count
38         if i < 0 || i >= m || j < 0 || j >= n || visited[i][j] || matrix[i][j] < pre
39         {
40             return
41         }
42         visited[i][j] = true
43         dfs(&matrix, &visited, matrix[i][j], i + 1, j)
44         dfs(&matrix, &visited, matrix[i][j], i - 1, j)
45         dfs(&matrix, &visited, matrix[i][j], i, j + 1)
46         dfs(&matrix, &visited, matrix[i][j], i, j - 1)
47     }
48 }

---

3572ms

 1 class Solution {
 2     var reachable: [[(Bool, Bool)]] = []
 3     
 4     func pacificAtlantic(_ matrix: [[Int]]) -> [[Int]] {
 5         guard !matrix.isEmpty && !matrix[0].isEmpty else { return [] }
 6         let n = matrix.count
 7         let m = matrix[0].count
 8         reachable  = Array(repeating: Array(repeating: (false, false), count: m), count: n)
 9         var visited: [[Bool]] = Array(repeating: Array(repeating: false, count: m), count: n)
10         
11         for i in 0 ..< n {
12             for j in 0 ..< m {
13                 dfs(matrix, i, j, &visited)
14             }
15         }
16         
17         var res: [[Int]] = []
18         
19         for i in 0 ..< n {
20             for j in 0 ..< m {
21                 if reachable[i][j].0 && reachable[i][j].1 {
22                     res.append([i, j])
23                 }
24             }
25         }
26         return res
27     }
28     
29     private func dfs(_ matrix: [[Int]], _ x: Int, _ y: Int, _ visited: inout [[Bool]]) -> (Bool, Bool) {
30         let n = matrix.count
31         let m = matrix[0].count
32         if reachable[x][y].0 && reachable[x][y].1 {
33             return (true, true)
34         }
35         visited[x][y] = true
36         let dirs: [(Int, Int)] = [(-1, 0), (0, -1), (1, 0), (0, 1)]
37         for dir in dirs {
38             let newX = x + dir.0
39             let newY = y + dir.1
40             if newX >= 0 && newX < n && newY >= 0 && newY < m && matrix[newX][newY] > matrix[x][y] {
41                 reachable[x][y].0 = reachable[x][y].0 || false
42                 reachable[x][y].1 = reachable[x][y].1 || false
43             } else if newX < 0 || newX >= n || newY < 0 || newY >= m {
44                 reachable[x][y].0 = reachable[x][y].0 || (newX < 0 || newY < 0)
45                 reachable[x][y].1 = reachable[x][y].1 || (newX >= n || newY >= m)
46             } else if !visited[newX][newY] {
47                 let res = dfs(matrix, newX, newY, &visited)
48                 reachable[x][y].0 = reachable[x][y].0 || res.0
49                 reachable[x][y].1 = reachable[x][y].1 || res.1
50             }
51         }
52         visited[x][y] = false
53         return reachable[x][y]
54     }
55 }