# 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  *          *   *   *   *   * AtlanticReturn:[[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) 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]: returncanReach[x][y] = True for each in direction: row = x + each col = y + each if row >= 0 and row < colLen and col >= 0 and col < rowLen and matrix[x][y] <= matrix[row][col]: dfs(canReach, row, col) returnfor 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 ansclass 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) 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, j + dir 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)`