# 417. Pacific Atlantic Water Flow 太平洋大西洋水流问题

## # 题目描述

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:

Return:

[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).


## # 解题方法

### # DFS

Python代码如下：

class Solution(object):
def pacificAtlantic(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[List[int]]
"""
if not matrix or not matrix[0]: return []
m, n = len(matrix), len(matrix[0])
p_visited = [[False] * n for _ in range(m)]
a_visited = [[False] * n for _ in range(m)]
for i in range(m):
self.dfs(p_visited, matrix, m, n, i, 0)
self.dfs(a_visited, matrix, m, n, i, n -1)
for j in range(n):
self.dfs(p_visited, matrix, m, n, 0, j)
self.dfs(a_visited, matrix, m, n, m - 1, j)
res = []
for i in range(m):
for j in range(n):
if p_visited[i][j] and a_visited[i][j]:
res.append([i, j])
return res

def dfs(self, visited, matrix, m, n, i, j):
visited[i][j] = True
directions = [(-1, 0), (1, 0), (0, 1), (0, -1)]
for dire in directions:
x, y = i + dire[0], j + dire[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(visited, matrix, m, n, x, y)


C++代码如下：

class Solution {
public:
vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
if (matrix.empty() || matrix[0].empty()) return {};
const int M = matrix.size();
const int N = matrix[0].size();
vector<vector<bool>> p_visited(M, vector<bool>(N));
vector<vector<bool>> a_visited(M, vector<bool>(N));
for (int i = 0; i < M; ++i) {
dfs(matrix, p_visited, i, 0);
dfs(matrix, a_visited, i, N - 1);
}
for (int j = 0; j < N; ++j) {
dfs(matrix, p_visited, 0, j);
dfs(matrix, a_visited, M - 1, j);
}
vector<pair<int, int>> res;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (p_visited[i][j] && a_visited[i][j]) {
res.push_back({i, j});
}
}
}
return res;
}
void dfs(vector<vector<int>>& matrix, vector<vector<bool>>& visited, int i, int j) {
const int M = matrix.size();
const int N = matrix[0].size();
visited[i][j] = true;
vector<pair<int, int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
for (auto d : dirs) {
int nx = i + d.first;
int ny = j + d.second;
if (nx >= 0 && nx < M && ny >= 0 && ny < N && !visited[nx][ny] && matrix[nx][ny] >= matrix[i][j]) {
dfs(matrix, visited, nx, ny);
}
}
}
};
`

https://leetcode.com/problems/pacific-atlantic-water-flow/discuss/90739/Python-DFS-bests-85.-Tips-for-all-DFS-in-matrix-question./181815open in new window

## # 日期

2018 年 10 月 1 日 —— 欢度国庆！