太平洋大西洋水流问题

/*
 * Solution in Java using Streams
 * This implementation is similar to the regular Java solution, but makes use of Java Stream API for readability.
 * The core logic is still based on DFS to mark reachable cells.
 */
 
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
 
public class Solution {
    public List<int[]> pacificAtlantic(int[][] heights) {
        int m = heights.length, n = heights[0].length;
        boolean[][] pacific = new boolean[m][n];
        boolean[][] atlantic = new boolean[m][n];
 
        IntStream.range(0, m).forEach(i -> {
            dfs(heights, pacific, i, 0);
            dfs(heights, atlantic, i, n - 1);
        });
        IntStream.range(0, n).forEach(j -> {
            dfs(heights, pacific, 0, j);
            dfs(heights, atlantic, m - 1, j);
        });
 
        return IntStream.range(0, m)
            .boxed()
            .flatMap(i -> IntStream.range(0, n)
                .filter(j -> pacific[i][j] && atlantic[i][j])
                .mapToObj(j -> new int[]{i, j}))
            .collect(Collectors.toList());
    }
 
    private void dfs(int[][] heights, boolean[][] ocean, int i, int j) {
        int m = heights.length, n = heights[0].length;
        ocean[i][j] = true;
        int[][] directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        for (int[] dir : directions) {
            int x = i + dir[0], y = j + dir[1];
            if (x >= 0 && x < m && y >= 0 && y < n && !ocean[x][y] && heights[x][y] >= heights[i][j]) {
                dfs(heights, ocean, x, y);
            }
        }
    }
}