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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:
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).
使用DFS来做
- public class Solution {
- private int m=0, n=0;
- public List<int[]> pacificAtlantic(int[][] matrix) {
- List<int[]> ans = new ArrayList<>();
- if(matrix==null || matrix.length==0 || matrix[0].length==0) return ans;
- m=matrix.length;
- n=matrix[0].length;
- boolean[][] p = new boolean[m][n];
- boolean[][] a = new boolean[m][n];
- for(int i=0; i<m; i++){
- helper(matrix, p, i, 0, Integer.MIN_VALUE);
- helper(matrix, a, i, n-1, Integer.MIN_VALUE);
- }
- for(int i=0; i<n; i++){
- helper(matrix, p, 0, i, Integer.MIN_VALUE);
- helper(matrix, a, m-1, i, Integer.MIN_VALUE);
- }
- for(int i=0; i<m; i++){
- for(int j=0; j<n; j++){
- if(p[i][j] && a[i][j]) ans.add(new int[]{i, j});
- }
- }
- return ans;
- }
- int[][] dir = new int[][]{{0,1}, {1,0}, {0,-1}, {-1,0}};
- private void helper(int[][] matrix, boolean[][] visited, int i, int j, int pre){
- if(i<0 || i>=m || j<0 || j>=n || visited[i][j] || matrix[i][j]<pre) return;
- visited[i][j] = true;
- for(int[] d : dir) helper(matrix, visited, i+d[0], j+d[1], matrix[i][j]);
- }
- }
- public class Solution {
- int[][]dir = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
- public List<int[]> pacificAtlantic(int[][] matrix) {
- List<int[]> res = new LinkedList<>();
- if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
- return res;
- }
- int n = matrix.length, m = matrix[0].length;
- //One visited map for each ocean
- boolean[][] pacific = new boolean[n][m];
- boolean[][] atlantic = new boolean[n][m];
- Queue<int[]> pQueue = new LinkedList<>();
- Queue<int[]> aQueue = new LinkedList<>();
- for(int i=0; i<n; i++){ //Vertical border
- pQueue.offer(new int[]{i, 0});
- aQueue.offer(new int[]{i, m-1});
- pacific[i][0] = true;
- atlantic[i][m-1] = true;
- }
- for(int i=0; i<m; i++){ //Horizontal border
- pQueue.offer(new int[]{0, i});
- aQueue.offer(new int[]{n-1, i});
- pacific[0][i] = true;
- atlantic[n-1][i] = true;
- }
- bfs(matrix, pQueue, pacific);
- bfs(matrix, aQueue, atlantic);
- for(int i=0; i<n; i++){
- for(int j=0; j<m; j++){
- if(pacific[i][j] && atlantic[i][j])
- res.add(new int[]{i,j});
- }
- }
- return res;
- }
- public void bfs(int[][]matrix, Queue<int[]> queue, boolean[][]visited){
- int n = matrix.length, m = matrix[0].length;
- while(!queue.isEmpty()){
- int[] cur = queue.poll();
- for(int[] d:dir){
- int x = cur[0]+d[0];
- int y = cur[1]+d[1];
- if(x<0 || x>=n || y<0 || y>=m || visited[x][y] || matrix[x][y] < matrix[cur[0]][cur[1]]){
- continue;
- }
- visited[x][y] = true;
- queue.offer(new int[]{x, y});
- }
- }
- }
- }
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