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 only 0, 1, or 2.
classSolution {publicintorangesRotting(int[][] grid) {int totalOranges =0;Queue<int[]> q =newLinkedList<>();for (int i =0; i <grid.length; i++)for (int j =0; j < grid[i].length; j++) {if (grid[i][j] ==2)q.add(newint[] { i, j });if (grid[i][j] ==1|| grid[i][j] ==2) totalOranges++; }int[][] dir = { { 1,0 }, { 0,1 }, { -1,0 }, { 0,-1 } };int minutes =0, rottenOranges =0;while (q.size() !=0) {int size =q.size();while (size-->0) {int[] pos =q.poll();for (int i =0; i <4; i++) {int x = pos[0] + dir[i][0];int y = pos[1] + dir[i][1];if (x >=0&& x <grid.length&& y >=0&& y < grid[x].length&& grid[x][y] ==1) { grid[x][y] =2;q.add(newint[] { x, y }); } } rottenOranges++; }if (q.size() !=0) minutes++; }return rottenOranges == totalOranges ? minutes :-1; }}