Swim in Rising Water

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 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

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].

思路：这题跟Path With Maximum Minimum Value 很相似，那题是path是最小，四周来的path取最大，这个是相反的，path是最大，然后四个path取最小；就是用pq每次poll最小的值，然后跟相邻的去比较，取两者最大，然后继续，最后到达n-1,m-1就是需要的最小的maxvalue。

class Solution {
    private class Node {
        public int x;
        public int y;
        public int value;
        public Node(int x, int y, int value) {
            this.x = x;
            this.y = y;
            this.value = value;
        }
    }
    
    public int swimInWater(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        PriorityQueue pq = new PriorityQueue((a, b) -> (a.value - b.value));
        int n = grid.length;
        int m = grid[0].length;
        boolean[][] visited = new boolean[n][m];
        pq.offer(new Node(0, 0, grid[0][0]));
        
        int[][] dirs = {
    {0,1},{0,-1},{-1,0},{1,0}};
        while(!pq.isEmpty()) {
            Node node = pq.poll();
            int x = node.x;
            int y = node.y;
            int value = node.value;
            if(visited[x][y]) {
                continue;
            }
            visited[x][y] = true;
            if(x == n - 1 && y == m - 1) {
                return value;
            }
            for(int[] dir: dirs) {
                int nx = x + dir[0];
                int ny = y + dir[1];
                if(0 <= nx && nx < n && 0 <= ny && ny < m && !visited[nx][ny]) {
                    pq.offer(new Node(nx, ny, Math.max(value, grid[nx][ny])));
                }
            }
        }
        return -1;
    }
}