标签：already exit return note evel when poll output count
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:
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.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.
标签：already exit return note evel when poll output count
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12169911.html
