【LeetCode】417. Pacific Atlantic Water Flow 解题报告(Python & C++)
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址: https://leetcode.com/problems/pacific-atlantic-water-flow/description/
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:
- 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).
题目大意
上面一条边和左边一条边代表的是太平洋,右边一条边和下边一条边代表的是大西洋。现在告诉你水往低处流,问哪些位置的水能同时流进太平洋和大西洋?
解题方法
直接DFS求解。一般来说DFS需要有固定的起点,但是对于这个题,四条边界的每个位置都算作起点。
使用两个二维数组,分别记录每个位置的点能不能到达太平洋和大西洋。然后对4条边界进行遍历,看这些以这些边为起点能不能所有的地方。注意了,因为是从边界向中间去寻找,所以,这个时候是新的点要比当前的点海拔高才行。
最坏情况下的时间复杂度是O((M+N)*MN),空间复杂度是O(MN)。
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);
}
}
}
};
参考资料:
日期
2018 年 10 月 1 日 —— 欢度国庆!
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