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

 1 class Solution {
 2 public:
 3     vector<int> direction{-1, 0, 1, 0, -1};
 4 // 主函数
 5     vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) {
 6         if (matrix.empty() || matrix[0].empty()) {
 7                     return {};
 8         }
 9         vector<vector<int>> ans;
10         int m = matrix.size(), n = matrix[0].size();
11         vector<vector<bool>> can_reach_p(m, vector<bool>(n, false));
12         vector<vector<bool>> can_reach_a(m, vector<bool>(n, false));
13         for (int i = 0; i < m; ++i) {
14             dfs(matrix, can_reach_p, i, 0);
15             dfs(matrix, can_reach_a, i, n - 1);
16         }
17         for (int i = 0; i < n; ++i) {
18             dfs(matrix, can_reach_p, 0, i);
19             dfs(matrix, can_reach_a, m - 1, i);
20         }
21         for (int i = 0; i < m; i++) {
22             for (int j = 0; j < n; ++j) {
23                 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
24                         ans.push_back(vector<int>{i, j});
25         }
26         }
27         }
28         return ans;
29 }
30 // 辅函数
31 void dfs(const vector<vector<int>>& matrix, vector<vector<bool>>& can_reach,int r, int c) {
32     if (can_reach[r][c]) {
33         return;
34     }
35     can_reach[r][c] = true;
36     int x, y;
37     for (int i = 0; i < 4; ++i) {
38         x = r + direction[i], y = c + direction[i+1];
39             if (x >= 0 && x < matrix.size() && y >= 0 && y < matrix[0].size() && matrix[r][c] <= matrix[x][y]) {
40         dfs(matrix, can_reach, x, y);
41     }
42     }
43     }
44 };

标签：417,matrix,int,reach,dfs,vector,流水,&&,leetcode
来源： https://www.cnblogs.com/libin123/p/14657116.html