题解 | #腐烂的苹果#

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param grid int整型vector<vector<>>
     * @return int整型
     */
    int rotApple(vector<vector<int> >& grid) {
        // write code here
        int n = grid.size(), m = grid[0].size();
        queue<pair<int, int>> q;
        int fresh = 0, minutes = 0;

        // 初始化队列和统计完好苹果
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (grid[i][j] == 2) {
                    q.push({i, j});
                } else if (grid[i][j] == 1) {
                    fresh++;
                }
            }
        }

        // 方向数组，用于表示上下左右四个方向
        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        // BFS
        while (!q.empty() && fresh > 0) {
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                auto [x, y] = q.front();
                q.pop();
                for (auto& dir : directions) {
                    int nx = x + dir.first, ny = y + dir.second;
                    if (nx >= 0 && nx < n && ny >= 0 && ny < m && grid[nx][ny] == 1) {
                        grid[nx][ny] = 2;
                        q.push({nx, ny});
                        fresh--;
                    }
                }
            }
            minutes++;
        }

        return fresh == 0 ? minutes : -1;
    }
};