994. Rotting Oranges/BFS模板題

題目描述

In a given grid, each cell can have one of three values:

• the value 0 representing an empty cell;
• the value 1 representing a fresh orange;
• the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) 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 instead.

 

Example 1:

Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4

Example 2:

Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation:  The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.

Example 3:

Input: [[0,2]]
Output: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.

 

Note:

1. 1 <= grid.length <= 10
2. 1 <= grid[0].length <= 10
3. grid[i][j] is only 0, 1, or 2.

解析

這裏不需要另外設一個inq數組來標記是否入過隊，只需要將入隊的位置的值設爲2即可。

struct node{
    int x,y;
};
class Solution {
public:
    int orangesRotting(vector<vector<int>>& grid) {
        int m=grid.size(),n=grid[0].size(),num=0;
        queue<node> q;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==2) q.push({i,j});
                if(grid[i][j]==1) num++;  //好橘子個數
            }
        }
        int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};  //增量數組     
        int ans=0;
        while(!q.empty()&&num>0){
            ans++;
            int cnt=q.size();
            while(cnt--){  //逐層的BFS
                node now=q.front();
                q.pop();
                for(int i=0;i<4;i++){
                    int newx=now.x+dx[i],newy=now.y+dy[i];
                    if(newx>=0&&newx<m&&newy>=0&&newy<n&&grid[newx][newy]==1){
                        grid[newx][newy]=2;num--;q.push({newx,newy});
                    }
                }
            }            
        }
        return num>0?-1:ans;  //如果q.empty()滿足了，跳出循環，但num>0說明無法全部腐爛
    }
};