LeetCode 1162. As Far from Land as Possible - Java - BFS

Given an N x N grid containing only values 0 and 1, where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized and return the distance.

The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|.

If no land or water exists in the grid, return -1.

Example 1:

Input: [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation: 
The cell (1, 1) is as far as possible from all the land with distance 2.

Example 2:

Input: [[1,0,0],[0,0,0],[0,0,0]]
Output: 4
Explanation: 
The cell (2, 2) is as far as possible from all the land with distance 4.

Note:

• 1 <= grid.length == grid[0].length <= 100
• grid[i][j] is 0 or 1

Java代碼

class Solution {
    private static final int WATER = 0;
    private static final int LAND = 1;
    private static final int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    public int maxDistance(int[][] grid) {
        int n = grid.length;
        Queue<int[]> queue = new LinkedList<>();
        // 將所有陸地入Queue
        for (int r = 0; r < n; ++r) {
            for (int c = 0; c < n; ++c) {
                if (grid[r][c] == LAND) {
                    queue.offer(new int[]{r, c});
                }
            }
        }
        if (queue.isEmpty() || queue.size() == n * n) {
            // 無陸地或者無海洋
            return -1;
        }
        // 當前格子，聲明在while外面是爲了保存最後一次出Queue的格子
        int[] cur = null;
        while (!queue.isEmpty()) {
            cur = queue.poll();
            for (int[] dir : dirs) {
                int r = cur[0] + dir[0];
                int c = cur[1] + dir[1];
                if (r < 0 || r >= n || c < 0 || c >= n) {
                    continue;
                }
                if (grid[r][c] == WATER) {
                    // [r, c]和當前位置相比距離加1
                    grid[r][c] = grid[cur[0]][cur[1]] + 1;
                    queue.offer(new int[]{r, c});
                }
            }
        }
        // 由於陸地本身是1，所以最後距離要減1
        return grid[cur[0]][cur[1]] - LAND;
    }
}