408算法题leetcode--第40天

994. 腐烂的橘子

题目地址 ：994. 腐烂的橘子 - 力扣（LeetCode）

题解思路：bfs

时间复杂度：O(mn)

空间复杂度：O(mn)

代码:

class Solution {
public:
    int dir[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};

    int orangesRotting(vector<vector<int>>& grid) {
        // bfs
        int m = grid.size(), n = grid[0].size();
        int fresh = 0;
        queue<pair<int, int>>q;  // 存储栏橘子的位置
        // 第0分钟
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    fresh++;
                } else if (grid[i][j] == 2){
                    q.push({i, j});
                }
            }
        }
        int ret = 0;
        while(!q.empty()){
            int size = q.size();
            bool flag = false;
            for(int i = 0; i < size; i++){
                auto [x, y] = q.front();
                q.pop();
                for(int i = 0; i < 4; i++){
                    int next_x = x + dir[i][0], next_y = y + dir[i][1];
                    if(next_x < 0 || next_x >= m || next_y < 0 || next_y >= n){
                        continue;
                    }
                    if(grid[next_x][next_y] == 1){
                        grid[next_x][next_y] = 2;
                        q.push({next_x, next_y});
                        fresh--;
                        flag = true;
                    }
                }
            }
            if(flag){
                ret++;
            }
        }
        return fresh ? -1 : ret;
    }
};