<p>There is an <code>m x n</code> rectangular island that borders both the <strong>Pacific Ocean</strong> and <strong>Atlantic Ocean</strong>. The <strong>Pacific Ocean</strong> touches the island&#39;s left and top edges, and the <strong>Atlantic Ocean</strong> touches the island&#39;s right and bottom edges.</p>

<p>The island is partitioned into a grid of square cells. You are given an <code>m x n</code> integer matrix <code>heights</code> where <code>heights[r][c]</code> represents the <strong>height above sea level</strong> of the cell at coordinate <code>(r, c)</code>.</p>

<p>The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell&#39;s height is <strong>less than or equal to</strong> the current cell&#39;s height. Water can flow from any cell adjacent to an ocean into the ocean.</p>

<p>Return <em>a <strong>2D list</strong> of grid coordinates </em><code>result</code><em> where </em><code>result[i] = [r<sub>i</sub>, c<sub>i</sub>]</code><em> denotes that rain water can flow from cell </em><code>(r<sub>i</sub>, c<sub>i</sub>)</code><em> to <strong>both</strong> the Pacific and Atlantic oceans</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/waterflow-grid.jpg" style="width: 400px; height: 400px;" />
<pre>
<strong>Input:</strong> heights = [[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]]
<strong>Output:</strong> [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]
<strong>Explanation:</strong> The following cells can flow to the Pacific and Atlantic oceans, as shown below:
[0,4]: [0,4] -&gt; Pacific Ocean 
&nbsp;      [0,4] -&gt; Atlantic Ocean
[1,3]: [1,3] -&gt; [0,3] -&gt; Pacific Ocean 
&nbsp;      [1,3] -&gt; [1,4] -&gt; Atlantic Ocean
[1,4]: [1,4] -&gt; [1,3] -&gt; [0,3] -&gt; Pacific Ocean 
&nbsp;      [1,4] -&gt; Atlantic Ocean
[2,2]: [2,2] -&gt; [1,2] -&gt; [0,2] -&gt; Pacific Ocean 
&nbsp;      [2,2] -&gt; [2,3] -&gt; [2,4] -&gt; Atlantic Ocean
[3,0]: [3,0] -&gt; Pacific Ocean 
&nbsp;      [3,0] -&gt; [4,0] -&gt; Atlantic Ocean
[3,1]: [3,1] -&gt; [3,0] -&gt; Pacific Ocean 
&nbsp;      [3,1] -&gt; [4,1] -&gt; Atlantic Ocean
[4,0]: [4,0] -&gt; Pacific Ocean 
       [4,0] -&gt; Atlantic Ocean
Note that there are other possible paths for these cells to flow to the Pacific and Atlantic oceans.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> heights = [[1]]
<strong>Output:</strong> [[0,0]]
<strong>Explanation:</strong> The water can flow from the only cell to the Pacific and Atlantic oceans.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == heights.length</code></li>
	<li><code>n == heights[r].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= heights[r][c] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    // Analysis
    // Method1: 反向来求，就是已知接触的点。 来反向求取的他们的令居。。然后推送
    // 否则，正向来求取。。因为存在相互通路的关系。。（a <-> b) 否则不好求取。
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        int n_rows = heights.size();
        int n_cols = heights[0].size();
        vector<vector<int>> pacific, atlantic;

        for (int i = 0; i < n_rows; i++) {
            vector<int> v;
            for (int j = 0; j < n_cols; j++) {
                if (i == 0 || j == 0) {
                    v.push_back(1);
                } else {
                    v.push_back(0);
                }
               
            }
            pacific.push_back(v);
        }
        traverse(heights, pacific, n_rows, n_cols);

        // atlantic
        for (int i = 0; i < n_rows; i++) {
            vector<int> v;
            for (int j = 0; j < n_cols; j++) {
                if (i == (n_rows-1) || j == (n_cols-1)) {
                    v.push_back(1);
                } else {
                    v.push_back(0);
                }
               
            }
            atlantic.push_back(v);
        }
        traverse(heights, atlantic, n_rows, n_cols);

        vector<vector<int>> ret;
        for (int i = 0; i < n_rows; i++) {
            for (int j = 0; j < n_cols; j++) {
                if (pacific[i][j] == 1 && atlantic[i][j] == 1) {
                    vector<int> v;
                    v.push_back(i);
                    v.push_back(j);
                    ret.push_back(v);
                }
            }
        }

        return ret;
    }

    void traverse(vector<vector<int>>& heights, vector<vector<int>> & ocean, int n_rows, int n_cols) {
        set<pair<int, int>> visited;
        queue<pair<int, int>> to_visited;
        for (int i = 0; i < n_rows; i++) {
            for (int j = 0; j < n_cols; j++) {
                if (ocean[i][j] == 1) {
                    to_visited.push({i,j});
                    visited.insert({i, j});
                }
            }
        }

        while (!to_visited.empty()) {
            pair<int, int> p = to_visited.front();
            to_visited.pop();
            int x = p.first;
            int y = p.second;

            vector<pair<int, int>> v;
            v.push_back({x-1,y});
            v.push_back({x+1,y});
            v.push_back({x,y-1});
            v.push_back({x,y+1});
            
            for (auto q : v) {
                if (q.first >= 0 && q.first < n_rows &&
                    q.second >= 0 && q.second < n_cols
                    && heights[x][y] <= heights[q.first][q.second]) {           
                    if (visited.find(q) == visited.end()) {
                        to_visited.push(q);
                        visited.insert(q);
                        ocean[q.first][q.second] = 1;
                    }  
                }
            }
        }
    }
};
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