[Swift]LeetCode778. 水位上升的泳池中游泳 | Swim in Rising Water

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On an N x N grid, each square grid[i][j]represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time , you are in grid location .
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point  until time .
When the depth of water is , we can swim anywhere inside the grid.
0(0, 0)(1, 1)33

Example 2:

Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation:
 0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.Note:

1. 2 <= N <= 50.
2. grid[i][j] is a permutation of [0, ..., N*N - 1].

---

在一個 N x N 的座標方格 grid 中，每個方格的值 grid[i][j] 表示在位置 (i,j) 的平臺高度。

如今開始下雨了。當時間爲 t 時，此時雨水致使水池中任意位置的水位爲 t 。你能夠從一個平臺遊向四周相鄰的任意一個平臺，可是前提是此時水位必須同時淹沒這兩個平臺。假定你能夠瞬間移動無限距離，也就是默認在方格內部遊動是不耗時的。固然，在你游泳的時候你必須待在座標方格里面。

你從座標方格的左上平臺 (0，0) 出發。最少耗時多久你才能到達座標方格的右下平臺 (N-1, N-1)？

示例 1:

輸入: [[0,2],[1,3]]
輸出: 3
解釋:
時間爲0時，你位於座標方格的位置爲 
此時你不能遊向任意方向，由於四個相鄰方向平臺的高度都大於當前時間爲 0 時的水位。

等時間到達 3 時，你才能夠遊向平臺 (1, 1). 由於此時的水位是 3，座標方格中的平臺沒有比水位 3 更高的，因此你能夠遊向座標方格中的任意位置
(0, 0)。

示例2:

輸入: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
輸入: 16
解釋:
 0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

最終的路線用加粗進行了標記。
咱們必須等到時間爲 16，此時才能保證平臺 (0, 0) 和 (4, 4) 是連通的

提示:

1. 2 <= N <= 50.
2. grid[i][j] 位於區間 [0, ..., N*N - 1] 內。

---

Runtime: 3588 ms

Memory Usage: 20.1 MB

 1 class Solution {
 2     var dirs:[[Int]] = [[0, -1],[-1, 0],[0, 1],[1, 0]]
 3     func swimInWater(_ grid: [[Int]]) -> Int {
 4         var grid = grid
 5         var n:Int = grid.count
 6         var dp:[[Int]] = [[Int]](repeating:[Int](repeating:Int.max,count:n),count:n)
 7         helper(&grid, 0, 0, grid[0][0], &dp)
 8         return dp[n - 1][n - 1]
 9     }
10     
11     func helper(_ grid:inout [[Int]],_ x:Int,_ y:Int,_ cur:Int,_ dp:inout [[Int]])
12     {
13         var n:Int = grid.count
14         if x < 0 || x >= n || y < 0 || y >= n || max(cur, grid[x][y]) >= dp[x][y]
15         {
16             return
17         }
18         dp[x][y] = max(cur, grid[x][y])
19         for dir in dirs
20         {
21             helper(&grid, x + dir[0], y + dir[1], dp[x][y], &dp)
22         }        
23     }
24 }