[Swift]LeetCode286. 墙和门 $ Walls and Gates

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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

---

您将得到一个由这三个可能值初始化的m x n 2d网格。

1. -1 - 墙或障碍物。
2. 0 - 门.
3. INF - 无限意味着一个空房间。我们使用值2^31-1=2147483647表示inf，因为您可以假定到门的距离小于2147483647。

   把每个空房间填满到最近的门的距离。如果不可能到达一个门，应该用inf填充。

把每个空房间填满到最近的门的距离。如果不可能到达一个门，应该用inf填充。

例如，给定二维网格：

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

运行函数后，二维网格应为：

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

---

Solution:BFS (Time Limit Exceeded)

class Solution {
    func wallsAndGates(_ rooms:inout [[Int]]) {
        var q:[(Int,Int)] = [(Int,Int)]()
        var dirs:[[Int]] = [[0, -1],[-1, 0],[0, 1],[1, 0]]
        for i in 0..<rooms.count
        {
            for j in 0..<rooms[i].count
            {
                if rooms[i][j] == 0
                {
                    q.append((i, j))
                }
            }
        }
        while(!q.isEmpty)
        {
            let i:Int = q.first!.0
            let j:Int = q.first!.1
            q.popLast()
            for k in 0..<dirs.count
            {
                let x:Int = i + dirs[k][0]
                let y:Int = j + dirs[k][1]
                if x < 0 || x >= rooms.count || y < 0 || y >= rooms[0].count || rooms[x][y] < rooms[i][j] + 1
                {
                    continue
                }
                rooms[x][y] = rooms[i][j] + 1
                q.append((x, y))
            }
        }
    }
}

---

Solution:DFS (严重超时产生错误)

class Solution {
    func wallsAndGates(_ rooms:inout [[Int]]) {
        for i in 0..<rooms.count
        {
            for j in 0..<rooms[i].count
            {
                if rooms[i][j] == 0
                {
                    dfs(&rooms, i, j, 0)
                }
            }
        }
    }
    
    func dfs(_ rooms:inout [[Int]],_ i:Int,_ j:Int,_ val:Int)
    {
        if i < 0 || i >= rooms.count || j < 0 || j >= rooms[i].count || rooms[i][j] < val
        {
            return
        }
        dfs(&rooms, i + 1, j, val + 1)
        dfs(&rooms, i - 1, j, val + 1)
        dfs(&rooms, i, j + 1, val + 1)
        dfs(&rooms, i, j - 1, val + 1)
    }
}

---

点击：Playground测试

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