题目链接

417. 太平洋大西洋水流问题

思路

问题可以转换成从四个边界出发，能和内部哪些点连通。

因为涉及到两个可达性问题，所以用两个 boolean 类型矩阵分别记录一个点到太平洋和大西洋的可达性。

代码

class Solution {
    public List<List<Integer>> pacificAtlantic(int[][] heights) {
        int m = heights.length;
        int n = heights[0].length;
        List<List<Integer>> result = new ArrayList<>();

        Queue<Pair<Integer, Integer>> queueLeftUp = new LinkedList<>();
        Queue<Pair<Integer, Integer>> queueRightDown = new LinkedList<>();
        boolean[][] accessLeftUp = new boolean[m][n];
        boolean[][] accessRightDown = new boolean[m][n];

        for(int i = 0; i < m; i++){
            queueLeftUp.add(new Pair<>(i, 0));
            accessLeftUp[i][0] = true;

            queueRightDown.add(new Pair<>(i, n - 1));
            accessRightDown[i][n - 1] = true;
        }
        for(int i = 0; i < n; i++){
            queueLeftUp.add(new Pair<>(0, i));
            accessLeftUp[0][i] = true;

            queueRightDown.add(new Pair<>(m - 1, i));
            accessRightDown[m - 1][i] = true;
        }

        bfs(queueLeftUp, accessLeftUp, heights, m, n);
        bfs(queueRightDown, accessRightDown, heights, m, n);

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

        return result;
    }

    void bfs(Queue<Pair<Integer, Integer>> queue, boolean[][] access, int[][] heights, int m, int n){
        boolean[][] visited = new boolean[m][n];

        while(!queue.isEmpty()){
            int[] dx = new int[]{1, 0, -1, 0};
            int[] dy = new int[]{0, 1, 0, -1};
            int x = queue.peek().getKey();
            int y = queue.remove().getValue();
            int height = heights[x][y];

            for(int i = 0; i < 4; i++){
                int nextX = x + dx[i];
                int nextY = y + dy[i];
                if(valid(nextX, nextY, visited, height, heights, m, n)){
                    access[nextX][nextY] = true;
                    queue.add(new Pair<>(nextX, nextY));
                    visited[nextX][nextY] = true;
                }
            }
        }
    }

    private boolean valid(int nextX, int nextY, boolean[][] visited, int height, int[][] heights, int m, int n) {
        return (
                0 <= nextX && nextX < m && 0 <= nextY && nextY < n &&
                        !visited[nextX][nextY] && height <= heights[nextX][nextY]
                );
    }
}