417. Pacific Atlantic Water Flow

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Problem

There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean. The Pacific Ocean touches the island’s left and top edges, and the Atlantic Ocean touches the island’s right and bottom edges.

The island is partitioned into a grid of square cells. You are given an m x n integer matrix heights where heights[r][c] represents the height above sea level of the cell at coordinate (r, c).

The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell’s height is less than or equal to the current cell’s height. Water can flow from any cell adjacent to an ocean into the ocean.

Return a 2D list of grid coordinates result where result[i] = [ri, ci] denotes that rain water can flow from cell (ri, ci) to both the Pacific and Atlantic oceans.

Example 1:

Input: heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
Output: [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]

Example 2:

Input: heights = [[2,1],[1,2]]
Output: [[0,0],[0,1],[1,0],[1,1]]

Code

class Solution {
    public List<List<Integer>> pacificAtlantic(int[][] heights) {
        List<List<Integer>> res = new ArrayList<>();
        if(heights == null || heights.length == 0) return res;

        int m = heights.length;
        int n = heights[0].length;

        boolean[][] pacific = new boolean[m][n];
        boolean[][] atlantic = new boolean[m][n];

        for(int i = 0; i < m; i++){
            dfs(heights, pacific, Integer.MIN_VALUE, i, 0);
            dfs(heights, atlantic, Integer.MIN_VALUE, i, n - 1);
        }

        for(int j = 0; j < n; j++){
            dfs(heights, pacific, Integer.MIN_VALUE, 0, j);
            dfs(heights, atlantic, Integer.MIN_VALUE, m - 1, j);
        }

        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(pacific[i][j] && atlantic[i][j]){
                    res.add(Arrays.asList(i, j));
                }
            }
        }

        return res;
    }

    private void dfs(int[][] heights, boolean[][] visited, int pre, int i, int j){
        if(i < 0 || j < 0 || i >= heights.length || j >= heights[0].length) return;
        if(visited[i][j] == true) return;
        // 当前值小于之前的值，表示水被拦住了
        if(heights[i][j] < pre) return;

        visited[i][j] = true;
        dfs(heights, visited, heights[i][j], i + 1, j);
        dfs(heights, visited, heights[i][j], i - 1, j);
        dfs(heights, visited, heights[i][j], i, j - 1);
        dfs(heights, visited, heights[i][j], i, j + 1);
    }
}

class Solution {
    int[][] DIRECTIONS = new int[][]{{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
    int m;
    int n;
    int[][] matrix;
    
    public List<List<Integer>> pacificAtlantic(int[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0) {
            return new ArrayList<>();
        }

        m = matrix.length;
        n = matrix[0].length;
        this.matrix = matrix;
        
        Queue<int[]> pQueue = new LinkedList<>();
        Queue<int[]> aQueue = new LinkedList<>();
        
        for (int i = 0; i < m; i++) {
            pQueue.offer(new int[]{i, 0});
            aQueue.offer(new int[]{i, n - 1});
        }
        
        for (int i = 0; i < n; i++) {
            pQueue.offer(new int[]{0, i});
            aQueue.offer(new int[]{m - 1, i});
        }
        
        boolean[][] pRes = bfs(pQueue);
        boolean[][] aRes = bfs(aQueue);
        
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (pRes[i][j] && aRes[i][j]) {
                    res.add(List.of(i, j));
                }
            }
        }
        
        return res;
    }
    
    private boolean[][] bfs(Queue<int[]> queue) {
        boolean[][] res = new boolean[m][n];
        
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            res[cell[0]][cell[1]] = true;
            
            for (int[] dir : DIRECTIONS) {
                int newRow = cell[0] + dir[0];
                int newCol = cell[1] + dir[1];

                if (newRow < 0 || newRow >= m || newCol < 0 || newCol >= n) {
                    continue;
                }

                if (res[newRow][newCol]) {
                    continue;
                }

                if (matrix[newRow][newCol] < matrix[cell[0]][cell[1]]) {
                    continue;
                }

                queue.offer(new int[]{newRow, newCol});
            }
        }
        
        return res;
    }
}