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:
The order of returned grid coordinates does not matter.
publicclassSolution {// Use 2 visited matrices// Now in our dfs we are only marking visited// if the current height >= previous heightpublicList<List<Integer>> pacificAtlantic(int[][] matrix) {List<List<Integer>> res =newLinkedList<>();if (matrix ==null||matrix.length==0|| matrix[0].length==0) {return res; }int n =matrix.length, m = matrix[0].length;boolean[][] pacific =newboolean[n][m];boolean[][] atlantic =newboolean[n][m];for (int i =0; i < n; i++) {dfs(matrix, pacific,Integer.MIN_VALUE, i,0);dfs(matrix, atlantic,Integer.MIN_VALUE, i, m -1); }for (int i =0; i < m; i++) {dfs(matrix, pacific,Integer.MIN_VALUE,0, i);dfs(matrix, atlantic,Integer.MIN_VALUE, n -1, i); }for (int i =0; i < n; i++)for (int j =0; j < m; j++)if (pacific[i][j] && atlantic[i][j]) {List<Integer> point =newArrayList<>();point.add(i);point.add(j);res.add(point); }return res; }int[][] dir =newint[][] { { 0,1 }, { 0,-1 }, { 1,0 }, { -1,0 } };publicvoiddfs(int[][] matrix,boolean[][] visited,int height,int x,int y) {int n =matrix.length, m = matrix[0].length;if (x <0|| x >= n || y <0|| y >= m || visited[x][y] || matrix[x][y] < height)return; visited[x][y] =true;for (int[] d : dir)dfs(matrix, visited, matrix[x][y], x + d[0], y + d[1]); }}