leetcode 417 太平洋大西洋流水问题

 类似于前面那道题，可以每个节点搜但是复杂度太高，还是从外到内

class Solution {
public:
    vector<int> direction{-1, 0, 1, 0, -1};
// 主函数
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) {
                    return {};
        }
        vector<vector<int>> ans;
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<bool>> can_reach_p(m, vector<bool>(n, false));
        vector<vector<bool>> can_reach_a(m, vector<bool>(n, false));
        for (int i = 0; i < m; ++i) {
            dfs(matrix, can_reach_p, i, 0);
            dfs(matrix, can_reach_a, i, n - 1);
        }
        for (int i = 0; i < n; ++i) {
            dfs(matrix, can_reach_p, 0, i);
            dfs(matrix, can_reach_a, m - 1, i);
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; ++j) {
                if (can_reach_p[i][j] && can_reach_a[i][j]) {  //既能流到大西洋也能流到太平洋
                        ans.push_back(vector<int>{i, j});
        }
        }
        }
        return ans;
}
// 辅函数
void dfs(const vector<vector<int>>& matrix, vector<vector<bool>>& can_reach,int r, int c) {
    if (can_reach[r][c]) {
        return;
    }
    can_reach[r][c] = true;
    int x, y;
    for (int i = 0; i < 4; ++i) {
        x = r + direction[i], y = c + direction[i+1];
            if (x >= 0 && x < matrix.size() && y >= 0 && y < matrix[0].size() && matrix[r][c] <= matrix[x][y]) {
        dfs(matrix, can_reach, x, y);
    }
    }
    }
};