class Solution {
public:
int orangesRotting(vector<vector<int>>& grid){
int move[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
vector<int> x;
vector<int> y;
int book[grid.size()][grid[0].size()];
memset(book,0,sizeof(book));
int minute = 0;
int p = 0;
while(1){
int flag = 0;
//记录下每一次新的腐烂的橘子的位置
for(int i = 0;i < grid.size();i++){
for(int j = 0;j < grid[0].size();j++){
if(grid[i][j] == 2 && book[i][j] == 0){
book[i][j] = 1;
x.push_back(i);
y.push_back(j);
flag++;
}
}
}
//如果没有新的腐烂的橘子,退出循环
if(flag == 0){
break;
}
//以新的腐烂的橘子味中心,继续向上下左右腐烂其他的橘子
for(int i = p;i < x.size();i++){
for(int j = 0;j < 4;j++){
int xx = x[i] + move[j][0];
int yy = y[i] + move[j][1];
if(xx < 0 || yy <0 || xx == grid.size() || yy == grid[0].size()){
continue;
}
if(grid[xx][yy] == 0) continue;
grid[xx][yy] = 2;
}
}
//记录下p的值,节省下次循环的时间
p = x.size();
//时间加一
minute++;
}
//还有没腐烂的橘子,返回-1
for(int i = 0;i < grid.size();i++){
for(int j = 0;j < grid[0].size();j++){
if(grid[i][j] == 1){
return -1;
}
}
}
//等于1是会出现grid=【【0,2】】的情况,等于0是会出现grid=【【0】】的情况
if(minute == 1 || minute == 0) return 0;
//因为第一次就已经minute++了,所以结果减一
return minute-1;
}
};