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  *
          *   *   *   *   * Atlantic

Return:

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

這道題給了我們一個二維陣列，說是陣列的左邊和上邊是太平洋，右邊和下邊是大西洋，假設水能從高處向低處流，問我們所有能流向兩大洋的點的集合。剛開始我們沒有理解題意，以為加括號的點是一條路徑，連通兩大洋的，但是看來看去感覺也不太對，後來終於明白了，是每一個點單獨都路徑來通向兩大洋。那麼就是典型的搜尋問題，那麼我最開始想的是對於每個點來搜尋是否能到達邊緣，只不過搜尋的目標點不在是一個單點，而是所有的邊緣點，找這種思路寫出的程式碼無法通過OJ大資料集，那麼我們就要想辦法來優化我們的程式碼，優化的方法跟之前那道Surrounded Regions很類似，都是換一個方向考慮問題，既然從每個點像中間擴散會TLE，那麼我們從邊緣當作起點開始遍歷搜尋，然後標記能到達的點位true，分別標記出pacific和atlantic能到達的點，那麼最終能返回的點就是二者均為true的點。我們可以先用DFS來遍歷二維陣列，參見程式碼如下：

s

解法一：

class Solution {
public:
    vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) return {};
        vector<pair<int, int>> res;
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<bool>> pacific(m, vector<bool>(n, false));
        vector<vector<bool>> atlantic(m, vector<bool>(n, false));
        for (int i = 0; i < m; ++i) {
            dfs(matrix, pacific, INT_MIN, i, 0);
            dfs(matrix, atlantic, INT_MIN, i, n - 1);
        }
        for (int i = 0; i < n; ++i) {
            dfs(matrix, pacific, INT_MIN, 0, i);
            dfs(matrix, atlantic, INT_MIN, m - 1, i);
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (pacific[i][j] && atlantic[i][j]) {
                    res.push_back({i, j});
                }
            }
        }
        return res;
    }
    void dfs(vector<vector<int>>& matrix, vector<vector<bool>>& visited, int pre, int i, int j) {
        int m = matrix.size(), n = matrix[0].size();
        if (i < 0 || i >= m || j < 0 || j >= n || visited[i][j] || matrix[i][j] < pre) return;
        visited[i][j] = true;
        dfs(matrix, visited, matrix[i][j], i + 1, j);
        dfs(matrix, visited, matrix[i][j], i - 1, j);
        dfs(matrix, visited, matrix[i][j], i, j + 1);
        dfs(matrix, visited, matrix[i][j], i, j - 1);
    }
};

那麼BFS的解法也可以做，用queue來輔助，開始把邊上的點分別存入queue中，然後對應的map標記true，然後開始BFS遍歷，遍歷結束後還是找pacific和atlantic均標記為true的點加入res中返回即可，參見程式碼如下：

解法二：

class Solution {
public:
    vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) return {};
        vector<pair<int, int>> res;
        int m = matrix.size(), n = matrix[0].size();
        queue<pair<int, int>> q1, q2;
        vector<vector<bool>> pacific(m, vector<bool>(n, false)), atlantic = pacific;
        for (int i = 0; i < m; ++i) {
            q1.push({i, 0}); 
            q2.push({i, n - 1});
            pacific[i][0] = true;
            atlantic[i][n - 1] = true;
        }
        for (int i = 0; i < n; ++i) {
            q1.push({0, i});
            q2.push({m - 1, i});
            pacific[0][i] = true;
            atlantic[m - 1][i] = true;
        }
        bfs(matrix, pacific, q1);
        bfs(matrix, atlantic, q2);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (pacific[i][j] && atlantic[i][j]) {
                    res.push_back({i, j});
                }
            }
        }
        return res;
    }
    void bfs(vector<vector<int>>& matrix, vector<vector<bool>>& visited, queue<pair<int, int>>& q) {
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
        while (!q.empty()) {
            auto t = q.front(); q.pop();
            for (auto dir : dirs) {
                int x = t.first + dir[0], y = t.second + dir[1];
                if (x < 0 || x >= m || y < 0 || y >= n || visited[x][y] || matrix[x][y] < matrix[t.first][t.second]) continue;
                visited[x][y] = true;
                q.push({x, y});
            }
        }
    }
};

參考資料：