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. Both m and n are less than 150. Example: 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 * * * * * * Atlantic Return: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
这题考点在于需要设置两个visited数组
Two Queue and add all the Pacific border to one queue; Atlantic border to another queue.
Keep a visited matrix for each queue. In the end, add the cell visited by two queue to the result.
BFS: Water flood from ocean to the cell. Since water can only flow from high/equal cell to low cell, add the neighboor cell with height larger or equal to current cell to the queue and mark as visited.（逆流而上）
Solution 1: 我自己的DFS (beat 89%)
1 public class Solution { 2 int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; 3 public List<int[]> pacificAtlantic(int[][] matrix) { 4 List<int[]> res = new ArrayList<int[]>(); 5 if (matrix==null || matrix.length==0 || matrix[0].length==0) return res; 6 int n = matrix.length, m = matrix[0].length; 7 boolean[][] pVisited = new boolean[n][m]; 8 boolean[][] aVisited = new boolean[n][m]; 9 for (int i=0; i<n; i++) { 10 //pacific 11 dfs(matrix, i, 0, pVisited); 12 //atlatic 13 dfs(matrix, i, m-1, aVisited); 14 } 15 16 for (int j=0; j<m; j++) { 17 //pacific 18 dfs(matrix, 0, j, pVisited); 19 //atlatic 20 dfs(matrix, n-1, j, aVisited); 21 } 22 23 for (int i=0; i<n; i++) { 24 for (int j=0; j<m; j++) { 25 if (pVisited[i][j] && aVisited[i][j]) 26 res.add(new int[]{i, j}); 27 } 28 } 29 return res; 30 } 31 32 public void dfs(int[][] matrix, int i, int j, boolean[][] visited) { 33 int n = matrix.length, m = matrix[0].length; 34 visited[i][j] = true; 35 for (int[] dir : directions) { 36 int row = dir[0] + i; 37 int col = dir[1] + j; 38 if (row>=0 && row<n && col>=0 && col<m && !visited[row][col] && matrix[i][j]<=matrix[row][col]) 39 dfs(matrix, row, col, visited); 40 } 41 } 42 }
Solution 2: BFS， refer to https://discuss.leetcode.com/topic/62379/java-bfs-dfs-from-ocean/2
1 public class Solution { 2 int[][]dir = new int[][]{{1,0},{-1,0},{0,1},{0,-1}}; 3 public List<int[]> pacificAtlantic(int[][] matrix) { 4 List<int[]> res = new LinkedList<>(); 5 if(matrix == null || matrix.length == 0 || matrix[0].length == 0){ 6 return res; 7 } 8 int n = matrix.length, m = matrix[0].length; 9 //One visited map for each ocean 10 boolean[][] pacific = new boolean[n][m]; 11 boolean[][] atlantic = new boolean[n][m]; 12 Queue<int[]> pQueue = new LinkedList<>(); 13 Queue<int[]> aQueue = new LinkedList<>(); 14 for(int i=0; i<n; i++){ //Vertical border 15 pQueue.offer(new int[]{i, 0}); 16 aQueue.offer(new int[]{i, m-1}); 17 pacific[i][0] = true; 18 atlantic[i][m-1] = true; 19 } 20 for(int i=0; i<m; i++){ //Horizontal border 21 pQueue.offer(new int[]{0, i}); 22 aQueue.offer(new int[]{n-1, i}); 23 pacific[0][i] = true; 24 atlantic[n-1][i] = true; 25 } 26 bfs(matrix, pQueue, pacific); 27 bfs(matrix, aQueue, atlantic); 28 for(int i=0; i<n; i++){ 29 for(int j=0; j<m; j++){ 30 if(pacific[i][j] && atlantic[i][j]) 31 res.add(new int[]{i,j}); 32 } 33 } 34 return res; 35 } 36 public void bfs(int[][]matrix, Queue<int[]> queue, boolean[][]visited){ 37 int n = matrix.length, m = matrix[0].length; 38 while(!queue.isEmpty()){ 39 int[] cur = queue.poll(); 40 for(int[] d:dir){ 41 int x = cur[0]+d[0]; 42 int y = cur[1]+d[1]; 43 if(x<0 || x>=n || y<0 || y>=m || visited[x][y] || matrix[x][y] > matrix[cur[0]][cur[1]]){ 44 continue; 45 } 46 visited[x][y] = true; 47 queue.offer(new int[]{x, y}); 48 } 49 } 50 } 51 }