Pacific Atlantic Water Flow

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).

分析：

与其从每个点开始找，不如从每个边上开始往上走，看能reach到那些地方。

一样的用DFS来解决。

知识点：

Initialise list的时候，如果简单的用list = [[False*numer]*number]]

只会create shallow copy， 里面的list都是同一个list。不可以这么写，debug半天没发现这个问题。

正确的写法是list = [[False for _ in range(number)] for _ in range(number)]

def pacificAtlantic(self, matrix):
 if not matrix: return []
 direction = [[-1,0], [0,1], [1,0], [0,-1]]
 colLen = len(matrix)
 rowLen = len(matrix[0])
 canReachP = [[False for _ in range(rowLen)] for _ in range(colLen)]
 canReachA = [[False for _ in range(rowLen)] for _ in range(colLen)]
 ans = []
def dfs(canReach, x, y):
 if canReach[x][y]:
 return
canReach[x][y] = True
 for each in direction:
 row = x + each[0]
 col = y + each[1]
 if row >= 0 and row < colLen and col >= 0 and col < rowLen and matrix[x][y] <= matrix[row][col]:
 dfs(canReach, row, col)
 return
for i in range(rowLen):
 dfs(canReachP, 0, i)
 dfs(canReachA, colLen-1, i)
for j in range(colLen):
 dfs(canReachP, j, 0)
 dfs(canReachA, j, rowLen-1)
for n in range(colLen):
 for m in range(rowLen):
 if canReachP[n][m] and canReachA[n][m]:
 ans.append([n,m])
 return ans
class Solution(object):
 def pacificAtlantic(self, matrix):
 “””
 :type matrix: List[List[int]]
 :rtype: List[List[int]]
 “””
 if not matrix: return []
 self.directions = [[-1,0], [0,1], [1,0], [0,-1]]
 colLen = len(matrix)
 rowLen = len(matrix[0])
 canReachP = [[False for _ in range(rowLen)] for _ in range(colLen)]
 canReachA = [[False for _ in range(rowLen)] for _ in range(colLen)]
 ans = []
 
 for j in range(colLen):
 self.dfs(matrix, j, 0, canReachP, colLen, rowLen)
 self.dfs(matrix, j, rowLen-1, canReachA, colLen, rowLen)
 
 for i in range(rowLen):
 self.dfs(matrix, 0, i, canReachP, colLen, rowLen)
 self.dfs(matrix, colLen-1, i, canReachA, colLen, rowLen)
 
 
 for n in range(colLen):
 for m in range(rowLen):
 if canReachP[n][m] and canReachA[n][m]:
 ans.append([n,m])
 return ans
 
 def dfs(self, matrix, i, j, visited, m, n):
 # when dfs called, meaning its caller already verified this point 
 visited[i][j] = True
 for dir in self.directions:
 x, y = i + dir[0], j + dir[1]
 if x < 0 or x >= m or y < 0 or y >= n or visited[x][y] or matrix[x][y] < matrix[i][j]:
 continue
 self.dfs(matrix, x, y, visited, m, n)