In a given grid, each cell can have one of three values:
- the value
0
representing an empty cell; - the value
1
representing a fresh orange; - the value
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 <= 10
1 <= grid[0].length <= 10
grid[i][j]
is only0
,1
, or2
.
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
class Solution { public int orangesRotting(int[][] grid) { if(grid == null || grid.length == 0) return 0; int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; int m = grid.length, n = grid[0].length; Queue<int[]> q = new LinkedList(); int fresh = 0; for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { if(grid[i][j] == 2) q.offer(new int[]{i, j}); else if(grid[i][j] == 1) fresh++; } } if(fresh == 0) return 0; int res = 0; System.out.print(q.size()); while(!q.isEmpty()) { res++; int si = q.size(); for(int i = 0; i < si; i++) { int[] cur = q.poll(); for(int[] dir : dirs) { int newi = cur[0] + dir[0]; int newj = cur[1] + dir[1]; if(newi >= 0 && newi < m && newj >= 0 && newj < n && grid[newi][newj] == 1) { grid[newi][newj] = 2; q.offer(new int[]{newi, newj}); fresh--; } } } } return fresh == 0 ? res - 1 : -1; } }
每一步扩展1,注意res要减1,因为只要多算了一次(如果只有rotten的也会进循环)