LeetCode 994. Rotting Oranges

原题链接在这里：https://leetcode.com/problems/rotting-oranges/

题目：

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. 1 <= grid.length <= 10
2. 1 <= grid[0].length <= 10
3. grid[i][j] is only 0, 1, or 2.

题解：

Iterate grid, for rotten orange, add it to the queue, for fresh orange, count++.

Perform BFS, when neibor is fresh, mark it as rotton and add to que, count--.

If eventually count == 0, then all rotton. return level.

Time Complexity: O(m * n). m = grid.length. n = grid[0].length.

Space: O(m * n).

AC Java:

 1 class Solution {
 2     public int orangesRotting(int[][] grid) {
 3         if(grid == null || grid.length == 0){
 4             return 0;
 5         }
 6
 7         int m = grid.length;
 8         int n = grid[0].length;
 9         LinkedList<int []> que = new LinkedList<>();
10         int cnt = 0;
11         for(int i = 0; i < m; i++){
12             for(int j = 0; j < n; j++){
13                 if(grid[i][j] == 2){
14                     que.add(new int[]{i, j});
15                 }else if(grid[i][j] == 1){
16                     cnt++;
17                 }
18             }
19         }
20
21         if(cnt == 0){
22             return 0;
23         }
24
25         int [][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
26         int level = -1;
27         while(!que.isEmpty()){
28             level++;
29             int size = que.size();
30             while(size-- > 0){
31                 int [] cur = que.poll();
32                 grid[cur[0]][cur[1]] = 2;
33
34                 for(int [] dir : dirs){
35                     int x = cur[0] + dir[0];
36                     int y = cur[1] + dir[1];
37                     if(x < 0 || x >= m || y < 0 || y >= n || grid[x][y] != 1){
38                         continue;
39                     }
40
41                     grid[x][y] = 2;
42                     cnt--;
43                     que.add(new int[]{x, y});
44                 }
45             }
46         }
47
48         return cnt == 0 ? level : -1;
49     }
50 }

类似Walls and Gates, Shortest Distance from All Buildings.

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12169911.html