# Pacific Atlantic Water Flow

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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).

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