Pacific Atlantic Water Flow

Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.

Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.

Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.

Note: 1.The order of returned grid coordinates does not matter. 2.Both m and n are less than 150.

Example:

 Given the following 5x5 matrix:

   Pacific ~   ~   ~   ~   ~ 
        ~  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  *
           *   *   *   *   * Atlantic

 Return:

 [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).

Tips:

O(MN)

分别从两个海往里走，如果走得到就在两个海中标记true，最后把两个海都标记了true的加入res。

Code：

dfs：

public class Solution {
    int[][]dir = new int[][]{{0,1},{0,-1},{1,0},{-1,0}};

    public List<int[]> pacificAtlantic(int[][] matrix) {
        List<int[]> res = new ArrayList<>();
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return res;
        }
        int m = matrix.length, n = matrix[0].length;
        boolean[][] p = new boolean[m][n];
        boolean[][] a = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            dfs(p, matrix, i, 0);
            dfs(a, matrix, i, n - 1);
        }
        for (int i = 0; i < n; i++) {
            dfs(p, matrix, 0, i);
            dfs(a, matrix, m - 1, i);
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (a[i][j] && p[i][j]) res.add(new int[] {i, j});
            }
        }
        return res;
    }

    private void dfs(boolean[][] visited, int[][] matrix, int i, int j) {
        int m = matrix.length, n = matrix[0].length;
        visited[i][j] = true;
        for (int[] d : dir) {
            int x = i + d[0];
            int y = j + d[1];
            if (x < 0 || y < 0 || x >= m || y >= n || visited[x][y] || matrix[x][y] < matrix[i][j]) continue;
            dfs(visited, matrix, x, y);
        }
    }
}

bfs：

public class Solution {
    int[][]dir = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
    public List<int[]> pacificAtlantic(int[][] matrix) {
        List<int[]> res = new LinkedList<>();
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return res;
        }
        int n = matrix.length, m = matrix[0].length;
        //One visited map for each ocean
        boolean[][] pacific = new boolean[n][m];
        boolean[][] atlantic = new boolean[n][m];
        Queue<int[]> pQueue = new LinkedList<>();
        Queue<int[]> aQueue = new LinkedList<>();
        for(int i=0; i<n; i++){ //Vertical border
            pQueue.offer(new int[]{i, 0});
            aQueue.offer(new int[]{i, m-1});
            pacific[i][0] = true;
            atlantic[i][m-1] = true;
        }
        for(int i=0; i<m; i++){ //Horizontal border
            pQueue.offer(new int[]{0, i});
            aQueue.offer(new int[]{n-1, i});
            pacific[0][i] = true;
            atlantic[n-1][i] = true;
        }
        bfs(matrix, pQueue, pacific);
        bfs(matrix, aQueue, atlantic);
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(pacific[i][j] && atlantic[i][j])
                    res.add(new int[]{i,j});
            }
        }
        return res;
    }
    public void bfs(int[][]matrix, Queue<int[]> queue, boolean[][]visited){
        int n = matrix.length, m = matrix[0].length;
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            for(int[] d:dir){
                int x = cur[0]+d[0];
                int y = cur[1]+d[1];
                if(x<0 || x>=n || y<0 || y>=m || visited[x][y] || matrix[x][y] < matrix[cur[0]][cur[1]]){
                    continue;
                }
                visited[x][y] = true;
                queue.offer(new int[]{x, y});
            } 
        }
    }
}