You are given an m x n grid where each cell can have one of three values:
0representing an empty cell,1representing a fresh orange, or2representing a rotten orange.
Every minute, any fresh orange that is 4-directionally adjacent 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.
Example 1:
Input: grid = [[2,1,1],[1,1,0],[0,1,1]] Output: 4
Example 2:
Input: grid = [[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: grid = [[0,2]] Output: 0 Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 10grid[i][j]is0,1, or2.
腐烂的橘子。
在给定的网格中，每个单元格可以有以下三个值之一：
值 0 代表空单元格；
值 1 代表新鲜橘子；
值 2 代表腐烂的橘子。
每分钟，任何与腐烂的橘子（在 4 个正方向上）相邻的新鲜橘子都会腐烂。返回直到单元格中没有新鲜橘子为止所必须经过的最小分钟数。如果不可能，返回 -1。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/rotting-oranges
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题目说了是求每个烂橘子四周围的情况，所以思路也是 flood fill 类型的 BFS。这个题跟一般的 BFS 的题目不太一样的地方在于，需要考虑会不会有一直有好橘子不腐烂的情况，所以需要算一下一共有多少个好橘子 count，每次腐烂了需要减去；和需要记录遍历次数 round，因为这是需要返回的参数。
先遍历一遍 input，得到好橘子的个数 count，也将所有的坏橘子的坐标加入 queue。从 queue 中弹出的时候，判断每个坐标的上下左右是否有好橘子，若有，则把他标记为坏橘子，再放回 queue。while 循环跳出的条件是 count == 0 或者 queue 为空。最后判断如果 count 不为 0 则返回 -1，说明一直有好橘子存在；否则则返回轮数 round。
时间O(mn)
空间O(mn)
Java BFS实现一
1 class Solution { 2 public int orangesRotting(int[][] grid) { 3 int M = grid.length; 4 int N = grid[0].length; 5 Queue<int[]> queue = new LinkedList<>(); 6 int count = 0; 7 for (int i = 0; i < M; i++) { 8 for (int j = 0; j < N; j++) { 9 if (grid[i][j] == 1) { 10 count++; 11 } else if (grid[i][j] == 2) { 12 queue.offer(new int[] { i, j }); 13 } 14 } 15 } 16 17 int round = 0; 18 while (count > 0 && !queue.isEmpty()) { 19 round++; 20 int n = queue.size(); 21 for (int i = 0; i < n; i++) { 22 int[] orange = queue.poll(); 23 int r = orange[0]; 24 int c = orange[1]; 25 if (r - 1 >= 0 && grid[r - 1][c] == 1) { 26 grid[r - 1][c] = 2; 27 count--; 28 queue.offer(new int[] { r - 1, c }); 29 } 30 if (r + 1 < M && grid[r + 1][c] == 1) { 31 grid[r + 1][c] = 2; 32 count--; 33 queue.offer(new int[] { r + 1, c }); 34 } 35 if (c - 1 >= 0 && grid[r][c - 1] == 1) { 36 grid[r][c - 1] = 2; 37 count--; 38 queue.offer(new int[] { r, c - 1 }); 39 } 40 if (c + 1 < N && grid[r][c + 1] == 1) { 41 grid[r][c + 1] = 2; 42 count--; 43 queue.offer(new int[] { r, c + 1 }); 44 } 45 } 46 } 47 if (count > 0) { 48 return -1; 49 } else { 50 return round; 51 } 52 } 53 }
Java BFS实现二，个人觉得这种实现对坐标的管理更容易记住
1 class Solution { 2 public int orangesRotting(int[][] grid) { 3 int m = grid.length; 4 int n = grid[0].length; 5 Queue<int[]> queue = new LinkedList<>(); 6 int count = 0; 7 for (int i = 0; i < m; i++) { 8 for (int j = 0; j < n; j++) { 9 if (grid[i][j] == 1) { 10 count++; 11 } else if (grid[i][j] == 2) { 12 queue.offer(new int[] { i, j }); 13 } 14 } 15 } 16 17 int[][] DIRS = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; 18 int round = 0; 19 while (count > 0 && !queue.isEmpty()) { 20 round++; 21 int size = queue.size(); 22 for (int i = 0; i < size; i++) { 23 int[] orange = queue.poll(); 24 int x = orange[0]; 25 int y = orange[1]; 26 for (int[] dir : DIRS) { 27 int newX = x + dir[0]; 28 int newY = y + dir[1]; 29 // 邻居节点在矩阵内 + 也是个烂橘子 30 if (newX >= 0 && newX < m && newY >= 0 && newY < n && grid[newX][newY] == 1) { 31 count--; 32 grid[newX][newY] = 2; 33 queue.offer(new int[] { newX, newY }); 34 } 35 } 36 } 37 } 38 if (count > 0) { 39 return -1; 40 } 41 return round; 42 } 43 }
JavaScript实现
1 /** 2 * @param {number[][]} grid 3 * @return {number} 4 */ 5 var orangesRotting = function (grid) { 6 let q = []; 7 let newFresh = 0; 8 let minutes = 0; 9 // push rotten oranges to the stack and count fresh oranges 10 for (let i = 0; i < grid.length; i++) { 11 for (let j = 0; j < grid[i].length; j++) { 12 if (grid[i][j] === 2) q.push([i, j]); 13 if (grid[i][j] === 1) newFresh++; 14 } 15 } 16 17 while (q.length && newFresh) { 18 let newQ = []; // queue for next minute 19 while (q.length) { 20 let badOrange = q.shift(); 21 let newRottens = infectOthers(grid, badOrange[0], badOrange[1], newQ); 22 newFresh -= newRottens; 23 } 24 minutes++; 25 q = newQ; 26 } 27 if (newFresh !== 0) return -1; 28 return minutes; 29 }; 30 31 // Infect surrounding oranges 32 // Return the number of newly infected oranges 33 var infectOthers = function (grid, i, j, q) { 34 let infected = 0; 35 if (i > 0 && grid[i - 1][j] === 1) { 36 grid[i - 1][j]++; 37 infected++; 38 q.push([i - 1, j]); 39 } 40 if (j > 0 && grid[i][j - 1] === 1) { 41 grid[i][j - 1]++; 42 infected++; 43 q.push([i, j - 1]); 44 } 45 if (i < grid.length - 1 && grid[i + 1][j] === 1) { 46 grid[i + 1][j]++; 47 infected++; 48 q.push([i + 1, j]); 49 } 50 if (j < grid[0].length - 1 && grid[i][j + 1] === 1) { 51 grid[i][j + 1]++; 52 infected++; 53 q.push([i, j + 1]); 54 } 55 return infected; 56 }