使陆地分离的最少天数

/*
题目思路：
Java Stream 不适合处理深度优先搜索（DFS）和复杂的矩阵操作。因此，在这个问题中，我们需要使用常规的Java来处理DFS和网格操作。
由于问题的复杂性和DFS的需求，我们使用Java常规代码来实现，而不使用Java Stream。
*/

public class Solution {
    private int[][] directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    
    public int minDays(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        if (countIslands(grid) != 1) return 0;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    grid[i][j] = 0;
                    if (countIslands(grid) != 1) return 1;
                    grid[i][j] = 1;
                }
            }
        }
        return 2;
    }
    
    private int countIslands(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        boolean[][] visited = new boolean[m][n];
        int islands = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    dfs(grid, visited, i, j);
                    islands++;
                }
            }
        }
        return islands;
    }
    
    private void dfs(int[][] grid, boolean[][] visited, int x, int y) {
        int m = grid.length;
        int n = grid[0].length;
        if (x < 0 || y < 0 || x >= m || y >= n || grid[x][y] == 0 || visited[x][y]) return;
        visited[x][y] = true;
        for (int[] dir : directions) {
            dfs(grid, visited, x + dir[0], y + dir[1]);
        }
    }
}