LeetCode

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).

上边界和左边界为Pacific，下边界和右边界为Atlantic，所有的洋流只能朝着小于等于其值的方向流动，问哪些可以既可以流入Pacific又可以流入Atlantic。

一个dfs，设置两个visit表示点（i，j）是否可以流入哪个大洋。时间复杂度O(nm)，空间复杂度O(nm)

class Solution {public:    vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {        vector<pair<int, int> > ans;        if (!matrix.size() || !matrix[0].size()) return ans;        int n = matrix.size(), m = matrix[0].size();                vector<vector<bool> > pacific(n, vector<bool>(m, false));        vector<vector<bool> > atlantic(n, vector<bool>(m, false));                for (int i = 0; i < n; ++i) {            dfs(matrix, pacific, i, 0, INT_MIN);            dfs(matrix, atlantic, i, m-1, INT_MIN);        }        for (int i = 0; i < m; ++i) {            dfs(matrix, pacific, 0, i, INT_MIN);            dfs(matrix, atlantic, n-1, i, INT_MIN);        }                for (int i = 0; i < n; ++i) {            for (int j = 0; j < m; ++j) {                if (pacific[i][j] && atlantic[i][j]) {                    ans.push_back(make_pair(i, j));                }            }        }        return ans;    }private:    int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};    void dfs(vector<vector<int>>& matrix, vector<vector<bool> >& vis, int x, int y, int height) {        int n = matrix.size(), m = matrix[0].size();        if (x < 0 || y < 0 || x >= n || y >= m || matrix[x][y] < height) return;        if (vis[x][y]) return;        vis[x][y] = true;           for (int i = 0; i < 4; ++i) {            dfs(matrix, vis, x + dir[i][0], y + dir[i][1], matrix[x][y]);        }        return;    }};