[LeetCode] 417. Pacific Atlantic Water Flow 太平洋大西洋水流

 

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數組，說是數組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。剛開始博主沒有理解題意，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，后來終於明白了，是每一個點單獨都有路徑來通向兩大洋。那么就是典型的搜索問題，我最開始想的是對於每個點都來搜索是否能到達邊緣，只不過搜索的目標點不再是一個單點，而是所有的邊緣點，照這種思路寫出的代碼無法通過 OJ 大數據集，那么就要想辦法來優化代碼，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，既然從每個點向中間擴散會 TLE，那么我們就把所有邊緣點當作起點開始遍歷搜索，然后標記能到達的點為 true，分別標記出 pacific 和 atlantic 能到達的點，那么最終能返回的點就是二者均為 true 的點。我們可以先用DFS來遍歷二維數組，參見代碼如下：

 

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);
    }
};

 

那么 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，然后對應的 map 標記 true，然后開始 BFS 遍歷，遍歷結束后還是找 pacific 和 atlantic 均標記為 true 的點加入結果 res 中返回即可，參見代碼如下：

 

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});
            }
        }
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/417

 

參考資料：

https://leetcode.com/problems/pacific-atlantic-water-flow/

https://leetcode.com/problems/pacific-atlantic-water-flow/discuss/90733/java-bfs-dfs-from-ocean

 

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