題目：417. 太平洋大西洋水流問題
思路
代碼
(1) MyMothed
// 符合條件的點 : 既可以到達左或上邊界，也可以到達右或下邊界；
class Solution {
private:int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};vector<vector<int>> result; void dfs(vector<vector<int>>& heights, vector<vector<bool>>& visited, int x, int y){int i, j;int next_x, next_y;int m = heights.size(), n = heights[0].size();for(i = 0; i < 4; i++){next_x = x + dir[i][0];next_y = y + dir[i][1];if(next_x < 0 || next_x >= m || next_y < 0 || next_y >= n){continue;}if(!visited[next_x][next_y] && heights[next_x][next_y] <= heights[x][y]){ visited[next_x][next_y] = true;dfs(heights, visited, next_x, next_y);}}}public:vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {int i, j;int m = heights.size(), n = heights[0].size();vector<vector<bool>> visited_P(m, vector<bool>(n, false)); // Pacific Oceanvector<vector<bool>> visited_A(m, vector<bool>(n, false)); // Atlantic Oceanresult.clear();for(i = 0; i < m; i++){for(j = 0; j < n; j++){heights[i][j] = -heights[i][j];}}// Pacifici = 0;for(j = 0; j < n; j++){if(!visited_P[i][j]){visited_P[i][j] = true;dfs(heights, visited_P, i, j);}}j = 0;for(i = 0; i < m; i++){if(!visited_P[i][j]){visited_P[i][j] = true;dfs(heights, visited_P, i, j);}}// Atlantici = m-1;for(j = 0; j < n; j++){if(!visited_A[i][j]){visited_A[i][j] = true;dfs(heights, visited_A, i, j);}}j = n-1;for(i = 0; i < m; i++){if(!visited_A[i][j]){visited_A[i][j] = true;dfs(heights, visited_A, i, j);}}cout << "P" << endl;for(i = 0; i < m; i++){for(j = 0; j < n; j++){cout << visited_P[i][j] << ", ";}cout << endl;}cout << "A" << endl;for(i = 0; i < m; i++){for(j = 0; j < n; j++){cout << visited_A[i][j] << ", ";}cout << endl;}// 能重合的地方就是想要的點for(i = 0; i < m; i++){for(j = 0; j < n; j++){if(visited_P[i][j] && visited_A[i][j]){result.push_back({i, j});}}}return result;}
};
把海拔翻轉一下，分別統計從太平洋能流經的地盤和大西洋能流經的地盤，這兩個地盤的重合部分就是題目所求；
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