LeetCode417

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.

这道题很有意思，有点类似于floodfill。在Solution里看到一个优雅的BFS和DFS写法，在这里记录一下。

BFS写法：用两个队列分别装入可达点，最后合并两个map的交集。

public class Solution {
    int[][]dir = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
    public List pacificAtlantic(int[][] matrix) {
        List 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 pQueue = new LinkedList<>();
        Queue aQueue = new LinkedList<>();
        for(int i=0; i 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});
            } 
        }
    }
}

DFS写法：相对简单一些，从边界开始搜索

public class Solution {
    public List pacificAtlantic(int[][] matrix) {
        List res = new LinkedList<>();
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return res;
        }
        int n = matrix.length, m = matrix[0].length;
        boolean[][]pacific = new boolean[n][m];
        boolean[][]atlantic = new boolean[n][m];
        for(int i=0; i=n || y<0 || y>=m || visited[x][y] || matrix[x][y] < height)
            return;
        visited[x][y] = true;
        for(int[]d:dir){
            dfs(matrix, visited, matrix[x][y], x+d[0], y+d[1]);
        }
    }
}