In a given grid, each cell can have one of three values:
0 representing an empty cell;1 representing a fresh orange;2 representing a rotten orange.Every minute, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.
Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1 instead.
Example 1:
Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
Example 2:
Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
Example 3:
Input: [[0,2]]
Output: 0
Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
Note:
1 <= grid.length <= 101 <= grid[0].length <= 10grid[i][j] is only 0, 1, or 2.class Solution { public int orangesRotting(int[][] grid) { if(grid == null || grid.length == 0) return 0; int rows = grid.length; int cols = grid[0].length; Queue<int[]> queue = new LinkedList<>(); int count_fresh = 0; //Put the position of all rotten oranges in queue //count the number of fresh oranges for(int i = 0 ; i < rows ; i++) { for(int j = 0 ; j < cols ; j++) { if(grid[i][j] == 2) { queue.offer(new int[]{i , j}); } else if(grid[i][j] == 1) { count_fresh++; } } } //if count of fresh oranges is zero --> return 0 if(count_fresh == 0) return 0; int count = 0; int[][] dirs = {{1,0},{-1,0},{0,1},{0,-1}}; //bfs starting from initially rotten oranges while(!queue.isEmpty()) { ++count; int size = queue.size(); for(int i = 0 ; i < size ; i++) { int[] point = queue.poll(); for(int dir[] : dirs) { int x = point[0] + dir[0]; int y = point[1] + dir[1]; //if x or y is out of bound //or the orange at (x , y) is already rotten //or the cell at (x , y) is empty //we do nothing if(x < 0 || y < 0 || x >= rows || y >= cols || grid[x][y] == 0 || grid[x][y] == 2) continue; //mark the orange at (x , y) as rotten grid[x][y] = 2; //put the new rotten orange at (x , y) in queue queue.offer(new int[]{x , y}); //decrease the count of fresh oranges by 1 count_fresh--; } } } return count_fresh == 0 ? count-1 : -1; } }
典型bfs
原文：https://www.cnblogs.com/wentiliangkaihua/p/13466936.html