题解 | #腐烂的苹果#

class Solution {
    int dx[4] = {1, -1, 0, 0};
    int dy[4] = {0, 0, 1, -1};
    int row, col;
  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) 
    {
        queue<pair<int, int>> q1;
        row = grid.size(), col = grid[0].size();
        for (int i = 0; i < row; i++) 
        {
            for (int j = 0; j < col; j++) 
            {
                if (grid[i][j] == 2) 
                {
                    q1.push({i, j});
                }
            }
        }
        int step  = 0;
        while (q1.size()) 
        {
            ++step;
            int sz = q1.size();
            while (sz--) 
            {
                auto [a, b] = q1.front();
                q1.pop();
                for (int i = 0; i < 4; i++) 
                {
                    int x = a + dx[i], y = b + dy[i];
                    if (x >= 0 && x < row && y >= 0 && y < col && grid[x][y] == 1) 
                    {
                        grid[x][y] = grid[a][b];
                        q1.push({x, y});
                    }
                }
            }
        }
        for (int i = 0; i < row; i++) 
        {
            for (int j = 0; j < col; j++) 
            {
                if(grid[i][j] == 1) return -1;
            }
        }
        return step-1;
    }
};