leetcode994
public class Solution
{
int row = ;
int column = ;
int FreshOrangeCount = ;
int RottenOrangeCount = ;
int Minute = ;
Queue<int[]> Q = new Queue<int[]>();
int[,] TagGrid;
public void BFS(int[][] grid)
{
var rottinglist = new List<int[]>(); while(Q.Count>)
{
while (Q.Count > )
{
var orange = Q.Dequeue(); int i = orange[];
int j = orange[]; int i_new = -;
int j_new = -; //合法范围:up,down,left,right
//未访问的:TagGrid[newi,newj]==0
//新鲜的(grid[i][j]==1)
i_new = i - ;
j_new = j; //up:i-1>=0
if (i_new >= && TagGrid[i_new, j_new] == && grid[i_new][j_new] == )
{
FreshOrangeCount--;
RottenOrangeCount++;
TagGrid[i_new, j_new] = ;//visited
grid[i_new][j_new] = ;//go bad
rottinglist.Add(new int[] { i_new, j_new });
} //down:i+1<=row-1
i_new = i + ;
j_new = j;
if (i_new <= row - && TagGrid[i_new, j_new] == && grid[i_new][j_new] == )
{
FreshOrangeCount--;
RottenOrangeCount++;
TagGrid[i_new, j_new] = ;//visited
grid[i_new][j_new] = ;//go bad
rottinglist.Add(new int[] { i_new, j_new });
} //left:j-1>=0
i_new = i;
j_new = j - ;
if (j_new >= && TagGrid[i_new, j_new] == && grid[i_new][j_new] == )
{
FreshOrangeCount--;
RottenOrangeCount++;
TagGrid[i_new, j_new] = ;//visited
grid[i_new][j_new] = ;//go bad
rottinglist.Add(new int[] { i_new, j_new });
} //right:j+1<=column-1
i_new = i;
j_new = j + ;
if (j_new <= column - && TagGrid[i_new, j_new] == && grid[i_new][j_new] == )
{
FreshOrangeCount--;
RottenOrangeCount++;
TagGrid[i_new, j_new] = ;//visited
grid[i_new][j_new] = ;//go bad
rottinglist.Add(new int[] { i_new, j_new });
} }
if (rottinglist.Any())
{
Minute++;
}
foreach (var l in rottinglist)
{
Q.Enqueue(l);
}
rottinglist.Clear();
}
} public int OrangesRotting(int[][] grid)
{
row = grid.Length;
column = grid[].Length;
TagGrid = new int[row, column]; for (var i = ; i < row; i++)
{
for (var j = ; j < column; j++)
{
TagGrid[i, j] = ;//unvisited
if (grid[i][j] == )
{
FreshOrangeCount++;
}
if (grid[i][j] == )
{
RottenOrangeCount++;
Q.Enqueue(new int[] { i, j });
}
}
} BFS(grid); if (FreshOrangeCount != )
{
return -;
}
else
{
return Minute;
}
}
}
leetcode994的更多相关文章
- [Swift]LeetCode994. 腐烂的橘子 | Rotting Oranges
In a given grid, each cell can have one of three values: the value 0 representing an empty cell; the ...
随机推荐
- 【转】non-blocking REST services with Spring MVC
堵塞Controller Controller为单例: 非线程安全: 堵塞方式: 1个request对应1个处理Thread: @RestController public class Process ...
- react事件中的事件对象和常见事件
不管是在原生的js还是vue中,所有的事件都有其事件对象,该事件对象event中包含着所有与事件相关的信息,在react中,所有的事件也有其事件对象,在触发DOM上的某个事件时,就会产生一个事件对象. ...
- react基础
上一篇文章主要是记录了自己是如何创建react项目的,今天则主要是总结一下react中的一个基础入门知识,包括数据定义和绑定.属性绑定.数组循环等等. 组件继承和挂载 当我们使用脚手架或者命令行创建一 ...
- Redis缓存机制
Redis介绍 Redis是一款内存高速缓存数据库: 数据模型为:key - value,非关系型数据库使用的存储数据的格式: 可持久化:将内存数据在写入之后按照一定格式存储在磁盘文件中,宕机.断电后 ...
- python图片和字符串的转换
有个业务,需要将图片压缩转化为64位编码上传到服务端. import json,requests,base64 #网上下载图片素材 r = requests.get("https://tim ...
- vSphere 查看FC HBA的WWNN和WWPN
# 查看WWN号
- Scrapy学习篇(四)之数据存储
上一篇中,我们简单的实现了toscrapy网页信息的爬取,并存储到mongo,本篇文章信息看看数据的存储.这一篇主要是实现信息的存储,我们以将信息保存到文件和mongo数据库为例,学习数据的存储,依然 ...
- [UE4]事件驱动的UI更新:事件调度器
事件调度器就是一个“事件中介”,可以被调用和被关注.
- 阿里云直播服务 sdk demo php
[php] <?php /** * Created by PhpStorm. * User: Administrator * Date: 2016/12/8 0008 * Time: 11:05 ...
- CRM 2016 请求"System.Security.Permissions.FilelOPermission,mscorlib,Version=4.0.0.0,Culture=neutral,PublicKeyToken=b77a5c561934e089"类型的权限已失败.
CRM 请求"System.Security.Permissions.FilelOPermission,mscorlib,Version=4.0.0.0,Culture=neutral,Pu ...