这篇博客探讨了LeetCode上的第994题，涉及如何处理矩阵中腐烂的橘子。题目要求计算最小时间，直到所有橘子都腐烂或证明无法全部腐烂。解决方案利用宽度优先搜索（BFS）策略，从腐烂的橘子开始，每一步检查并转化相邻的新鲜橘子，直到没有新鲜橘子剩余。文章还提供了具体的算法实现细节和例子解析。
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 ora
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