417. 太平洋大西洋水流问题（DFS + BFS + 方向数组）

Question

417. 太平洋大西洋水流问题

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/random-pick-index/
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Ideas

1、Answer( Java )

解法思路：DFS + 方向数组

⚡️ 方向数组 dirs

//对应上下左右四个方向
static int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

👍 逆向思维 反向搜索 对于一个点它能流动两边的大洋，那么反过来，两边大洋的水反着流就能达到这个点。

1. 找出所有从太平洋出发的水所能达到的点 canReachP
2. 找出所有从大西洋出发的水所能达到的点 canReachA
3. 重合的点即是要找的点

Code①（ DFS + 方向数组 ）

/**
 * @author Listen 1024
 * @description 417. 太平洋大西洋水流问题（DFS +  BFS + 方向数组）
 * @date 2022-04-27 8:18
 */
class Solution {
    static int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};//the direction array
    int m, n;
    int[][] heights;

    public List<List<Integer>> pacificAtlantic(int[][] heights) {
        ArrayList<List<Integer>> res = new ArrayList<>();
        m = heights.length;
        n = heights[0].length;
        this.heights = heights;
        boolean[][] canReachP = new boolean[m][n], canReachA = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            dfs(canReachP, i, 0);//begin with left boundary
            dfs(canReachA, i, n - 1);//begin with right boundary

        }
        for (int i = 0; i < n; i++) {
            dfs(canReachA, m - 1, i);
            dfs(canReachP, 0, i);

        }

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

        }
        return res;
    }

    private void dfs(boolean[][] canReach, int row, int col) {
        if (canReach[row][col]) {
            return;
        }
        canReach[row][col] = true;
        for (int[] dir : dirs) {
            int newRow = row + dir[0], newCol = col + dir[1];
            if (newRow >= 0 && newRow < m && newCol >= 0 && newCol < n && heights[newRow][newCol] >= heights[row][col]) {
                dfs(canReach, newRow, newCol);
            }
        }
    }
}

Code②（ DFS （ 方向数组思想 ））

/**
 * @author Listen 1024
 * @description 417. 太平洋大西洋水流问题（DFS +  BFS + 方向数组）
 * @date 2022-04-27 8:18
 */
class Solution {
    public List<List<Integer>> pacificAtlantic(int[][] heights) {
        ArrayList<List<Integer>> res = new ArrayList<>();
        int m = heights.length, n = heights[0].length;
        boolean[][] canReachP = new boolean[m][n], canReachA = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            dfs(heights, canReachP, i, 0);//begin with left boundary
            dfs(heights, canReachA, i, n - 1);//begin with right boundary

        }
        for (int i = 0; i < n; i++) {
            dfs(heights, canReachA, m - 1, i);
            dfs(heights, canReachP, 0, i);

        }

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

        }
        return res;
    }

    private void dfs(int[][] heights, boolean[][] canReach, int i, int j) {
        //return when the zone was reached
        if (canReach[i][j]) {
            return;
        }
        //set the current zone true
        canReach[i][j] = true;
        //Up
        if (i - 1 >= 0 && heights[i - 1][j] >= heights[i][j]) {
            dfs(heights, canReach, i - 1, j);
        }
        //Down
        if (i + 1 < heights.length && heights[i + 1][j] >= heights[i][j]) {
            dfs(heights, canReach, i + 1, j);
        }
        //Left
        if (j - 1 >= 0 && heights[i][j - 1] >= heights[i][j]) {
            dfs(heights, canReach, i, j - 1);
        }
        //Right
        if (j + 1 < heights[0].length && heights[i][j + 1] >= heights[i][j]) {
            dfs(heights, canReach, i, j + 1);
        }
    }
}

2、Answer( Java )

解法思路：BFS + 方向数组

Code（BFS + 方向数组 ）

/**
 * @author Listen 1024
 * @description 417. 太平洋大西洋水流问题（DFS +  BFS + 方向数组）
 * @date 2022-04-27 8:18
 */
 class Solution {
    static int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};//the direction array
    int m, n;
    int[][] heights;

    public List<List<Integer>> pacificAtlantic(int[][] heights) {
        ArrayList<List<Integer>> res = new ArrayList<>();
        m = heights.length;
        n = heights[0].length;
        this.heights = heights;
        boolean[][] canReachP = new boolean[m][n], canReachA = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            bfs(canReachP, i, 0);//begin with left boundary
            bfs(canReachA, i, n - 1);//begin with right boundary

        }
        for (int i = 0; i < n; i++) {
            bfs(canReachA, m - 1, i);
            bfs(canReachP, 0, i);

        }

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

        }
        return res;
    }

    private void bfs(boolean[][] canReach, int row, int col) {
        if (canReach[row][col]) {
            return;
        }
        canReach[row][col] = true;
        Queue<int[]> queue = new ArrayDeque<>();
        queue.offer(new int[]{row, col});
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            for (int[] dir : dirs) {
                int newRow = cell[0] + dir[0], newCol = cell[1] + dir[1];
                if (newRow >= 0 && newRow < m && newCol >= 0 && newCol < n && heights[newRow][newCol] >= heights[cell[0]][cell[1]] && !canReach[newRow][newCol]) {
                    canReach[newRow][newCol] = true;
                    queue.offer(new int[]{newRow, newCol});
                }
            }

        }

    }
}

//题解参考链接(如侵删)
https://leetcode.cn/problems/pacific-atlantic-water-flow/solution/tai-ping-yang-da-xi-yang-shui-liu-wen-ti-sjk3/
https://leetcode.cn/problems/pacific-atlantic-water-flow/solution/shui-wang-gao-chu-liu-by-xiaohu9527-xxsx/