417. Pacific Atlantic Water Flow

417. Pacific Atlantic Water Flow

题目链接：https://leetcode.com/problems...

思路是分别找到pacific和atlantic能够流到的地方，然后求两个地方的交集。找pacific和atlantic能流到的地方，就是这个matrix的遍历过程，可以用dfs或者bfs。复杂度没什么差，dfs写起来简单点。

public class Solution {
    public List<int[]> pacificAtlantic(int[][] matrix) {
        List<int[]> result = new ArrayList();
        if(matrix.length == 0 || matrix[0].length == 0) return result;
        
        m = matrix.length;
        n = matrix[0].length;
        boolean[][] pacific = new boolean[m][n];
        boolean[][] atlantic = new boolean[m][n];
        // dfs, for each position
        for(int i = 0; i < m; i++) {
            dfs(matrix, i, 0, pacific);
            dfs(matrix, i, n- 1, atlantic);
        }
        for(int j = 0; j < n; j++) {
            dfs(matrix, 0, j, pacific);
            dfs(matrix, m - 1, j, atlantic);
        }
        // find the intersection
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(pacific[i][j] && atlantic[i][j]) result.add(new int[] {i, j});
            }
        }
        return result;
    }
    
    int m, n;
    int[][] dirs = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
    private void dfs(int[][] matrix, int x, int y, boolean[][] visited) {
        if(visited[x][y]) return;
        visited[x][y] = true;
        for(int[] dir : dirs) {
            int nx = x + dir[0], ny = y + dir[1];
            if(nx >= 0 && nx < m && ny >= 0 && ny < n && matrix[x][y] <= matrix[nx][ny]) {
                dfs(matrix, nx, ny, visited);
            }
        }
    }
}