LeetCode 0417. Pacific Atlantic Water Flow太平洋大西洋水流问题【Medium】【Python】【DFS】

Problem

LeetCode

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).

问题

力扣

给定一个 m x n 的非负整数矩阵来表示一片大陆上各个单元格的高度。“太平洋”处于大陆的左边界和上边界，而“大西洋”处于大陆的右边界和下边界。

规定水流只能按照上、下、左、右四个方向流动，且只能从高到低或者在同等高度上流动。

请找出那些水流既可以流动到“太平洋”，又能流动到“大西洋”的陆地单元的坐标。

提示：

  1. 输出坐标的顺序不重要
  2. m 和 n 都小于150

示例：

给定下面的 5x5 矩阵:

  太平洋 ~   ~   ~   ~   ~ 
       ~  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  *
          *   *   *   *   * 大西洋

返回:

[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (上图中带括号的单元).

思路

DFS

从边界开始遍历，找到水可以流的位置。
因为是倒过来，所以应该是 [x,y] > [i,j] 才满足，即 [x,y] 应该比 [i,j] 高。

时间复杂度: O(m*n)
空间复杂度: O(m*n)

Python3代码
class Solution:
    def pacificAtlantic(self, matrix: List[List[int]]) -> List[List[int]]:
        if not matrix or not matrix[0]:
            return []

        m, n = len(matrix), len(matrix[0])
        p_visited = [[False] * n for x in range(m)]  # Pacific
        a_visited = [[False] * n for x in range(m)]  # Atlantic

        # left and right
        for i in range(m):
            self.dfs(p_visited, matrix, m, n, i, 0)
            self.dfs(a_visited, matrix, m, n, i, n - 1)

        # up and down
        for j in range(n):
            self.dfs(p_visited, matrix, m, n, 0, j)
            self.dfs(a_visited, matrix, m, n, m - 1, j)

        ans = []
        for i in range(m):
            for j in range(n):
                if p_visited[i][j] and a_visited[i][j]:
                    ans.append([i, j])
        return ans

    def dfs(self, visited, matrix, m, n, i, j):
        visited[i][j] = True
        directions = [(-1, 0), (0, 1), (1, 0), (0, -1)]
        for d in directions:
            x, y = i + d[0], j + d[1]
            if x < 0 or x >= m or y < 0 or y >= n or visited[x][y] or matrix[x][y] < matrix[i][j]:  # [x, y] < [i, j]
                continue
            self.dfs(visited, matrix, m, n, x, y)

代码地址

GitHub链接