Question
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:
- The order of returned grid coordinates does not matter.
- 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).
Solution
这道题的解题思路与Surrounded Regions类似。
已知在边缘上的点一定能到达Pacific或Atlantic，我们从边缘上的点开始遍历(BFS/DFS)，得到所有能到达Pacific或Atlantic的点，最后取交集就是答案。
DFS
1 class Solution: 2 def pacificAtlantic(self, matrix: List[List[int]]) -> List[List[int]]: 3 if not matrix or not matrix[0]: 4 return [] 5 m, n = len(matrix), len(matrix[0]) 6 result = [] 7 self.directions = [[0, -1], [0, 1], [-1, 0], [1, 0]] 8 pacific = [[False for i in range(n)] for j in range(m)] 9 atlantic = [[False for i in range(n)] for j in range(m)] 10 # DFS 11 for i in range(m): 12 self.dfs(matrix, pacific, matrix[i][0], i , 0) 13 self.dfs(matrix, atlantic, matrix[i][n - 1], i, n - 1) 14 for i in range(n): 15 self.dfs(matrix, pacific, matrix[0][i], 0, i) 16 self.dfs(matrix, atlantic, matrix[m - 1][i], m - 1, i) 17 18 for i in range(m): 19 for j in range(n): 20 if pacific[i][j] and atlantic[i][j]: 21 result.append([i, j]) 22 return result 23 24 def dfs(self, matrix: List[List[int]], visited: List[List[bool]], pre: int, i: int, j: int) -> None: 25 m, n = len(matrix), len(matrix[0]) 26 if i < 0 or i >= m or j < 0 or j >= n or visited[i][j] or matrix[i][j] < pre: 27 return 28 visited[i][j] = True 29 for direction in self.directions: 30 self.dfs(matrix, visited, matrix[i][j], i + direction[0], j + direction[1])
BFS
1 from collections import deque 2 class Solution: 3 def pacificAtlantic(self, matrix: List[List[int]]) -> List[List[int]]: 4 if not matrix or not matrix[0]: 5 return [] 6 m, n = len(matrix), len(matrix[0]) 7 result = [] 8 self.directions = [[0, -1], [0, 1], [-1, 0], [1, 0]] 9 pacific = [[False for i in range(n)] for j in range(m)] 10 atlantic = [[False for i in range(n)] for j in range(m)] 11 # BFS 12 for i in range(m): 13 self.bfs(matrix, pacific, i , 0) 14 self.bfs(matrix, atlantic, i, n - 1) 15 for i in range(n): 16 self.bfs(matrix, pacific, 0, i) 17 self.bfs(matrix, atlantic, m - 1, i) 18 19 for i in range(m): 20 for j in range(n): 21 if pacific[i][j] and atlantic[i][j]: 22 result.append([i, j]) 23 return result 24 25 def bfs(self, matrix: List[List[int]], visited: List[List[bool]], i: int, j: int) -> None: 26 # Avoid duplicate calculation 27 if visited[i][j]: 28 return 29 m, n = len(matrix), len(matrix[0]) 30 visited[i][j] = True 31 level = deque() 32 level.append((i, j)) 33 while level: 34 L = len(level) 35 for _ in range(L): 36 curr_i, curr_j = level.popleft() 37 for direction in self.directions: 38 next_i = curr_i + direction[0] 39 next_j = curr_j + direction[1] 40 if 0 <= next_i < m and 0 <= next_j < n and not visited[next_i][next_j] and matrix[next_i][next_j] >= matrix[curr_i][curr_j]: 41 visited[next_i][next_j] = True 42 level.append((next_i, next_j))