Pacific Atlantic Water Flow问题及解法

问题描述：

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:

1. The order of returned grid coordinates does not matter.
2. Both m and n are less than 150.

示例:

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).

问题分析：

本题主要求解共同的水流路径，我们可以利用位运算记录每种水流的路径，通过DFS将共同的路径保存起来。

过程详见代码：

class Solution {public:    vector<pair<int, int>> res;    vector<vector<int>> visited;    void dfs(vector<vector<int>>& matrix, int x, int y, int pre, int preval){        if (x < 0 || x >= matrix.size() || y < 0 || y >= matrix[0].size()                  || matrix[x][y] < pre || (visited[x][y] & preval) == preval)             return;        visited[x][y] |= preval;        if (visited[x][y] == 3) res.push_back({x, y});        dfs(matrix, x + 1, y, matrix[x][y], visited[x][y]); dfs(matrix, x - 1, y, matrix[x][y], visited[x][y]);        dfs(matrix, x, y + 1, matrix[x][y], visited[x][y]); dfs(matrix, x, y - 1, matrix[x][y], visited[x][y]);    }    vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {        if (matrix.empty()) return res;        int m = matrix.size(), n = matrix[0].size();        visited.resize(m, vector<int>(n, 0));        for (int i = 0; i < m; i++) {            dfs(matrix, i, 0, INT_MIN, 1);            dfs(matrix, i, n - 1, INT_MIN, 2);        }        for (int i = 0; i < n; i++) {            dfs(matrix, 0, i, INT_MIN, 1);            dfs(matrix, m - 1, i, INT_MIN, 2);        }        return res;    }};