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:這題跟之前的Dijkstra算法一模一樣,也就是找所有path裏面的最小path的最大值。就是用pq每次poll最小的值,然後跟相鄰的去比較,取兩者最大,然後繼續,最後到達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;
        }
    }
    
    private int[] dx = {0,0,-1,1};
    private int[] dy = {-1,1,0,0};
    
    public int swimInWater(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int n = grid.length;
        int m = grid[0].length;
        
        boolean[][] visited = new boolean[n][m];
        PriorityQueue<Node> pq = new PriorityQueue<Node>((a, b) -> a.value - b.value);
        pq.offer(new Node(0, 0, grid[0][0]));
        
        while(!pq.isEmpty()) {
            Node node = pq.poll();
            if(node.x == n - 1 && node.y == m - 1) {
                return node.value;
            }
            visited[node.x][node.y] = true;
            for(int k = 0; k < 4; k++) {
                int nx = node.x + dx[k];
                int ny = node.y + dy[k];
                if(0 <= nx && nx < n && 0 <= ny && ny < m && !visited[nx][ny]) {
                    pq.offer(new Node(nx, ny, Math.max(node.value, grid[nx][ny])));
                }
            }
        }
        return -1;
    }
}

 

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