0

在给定的网格中，每个单元格可以具有以下三个值之一：

值 0 表示一个空单元格；值 1 代表新鲜的橙子；值 2 代表一个烂橘子。每分钟，与腐烂的橙子相邻（4 个方向）的任何新鲜橙子都会腐烂。

返回在没有单元格有新鲜橙色之前必须经过的最小分钟数。如果这是不可能的，则返回 -1。

class Solution {
public:
    int orangesRotting(vector<vector<int>>& grid) {
        int m,n;
       m = grid.size();
        n = grid[0].size();

        int i, j, min = 0,flag=0,fresh=0;

        int r[4] = {-1,1,0,0};
        int c[4] = {0,0,-1,1};
        for(i=0;i<m;i++) {
            for(j=0;j<n;j++) {
                if(grid[i][j]==1) 
                    fresh++;
            }
        }
        queue< pair<int, int> >q;
        for(i=0;i<m;i++) {
            for(j=0;j<n;j++) {
                if(grid[i][j] == 2) {
                    q.push(make_pair(i,j));
                    flag=1;
                    break;
                }
            }
            if(flag==1)
                break;  
        }
        while(!q.empty()) {

            pair<int,int> p = q.front();
            int a = p.first;
            int b = p.second;
          int x=0;
            q.pop();
            for(i=0;i<4;i++) {
                for(j=0;j<4;j++) {
                    int rr = a + r[i];
                    int cc = b + c[j];
                    if(rr<0 || cc<0 || rr>=m || cc>=n || grid[rr][cc]==0 || grid[rr][cc] ==2) {
                        continue;
                    }
                    else if(grid[rr][cc] ==1) {
                         grid[rr][cc] =2;
                        q.push(make_pair(rr,cc));
                        fresh--;  
                        x++;
                    }     
                }
            }    
     if(x>0) min++;
        } 
     return fresh >0 ? -1:min; 
    }
};

输入：[[2,1,1],[1,1,0],[0,1,1]]

输出：3

预计：4

4

2 回答 2

1

编辑 1

您计算分钟的方法是错误的，每次腐烂的橙子至少将新鲜橙子变成烂橙子时，您都会增加分钟数。因为每分钟的结果分钟数也取决于你在烂橙上迭代的顺序，这是错误的。

橙子必须同时腐烂，您迭代到网格中的顺序必须不相关。

如果我在您的程序中添加每分钟网格的打印结果：

t = 0
211
110
011

t = 1
221
220
011

t = 2
221
220
021

t = 3
222
220
022

t = 3
222
220
022

t = 3
222
220
022

t = 3
222
220
022

与我的案例比较

编辑 2

您的提案中的更正方法可以是：

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int orangesRotting(vector<vector<int>>& grid) {
  int m,n;
  m = grid.size();
  n = grid[0].size();

  int i, j, min = 0,flag=0,fresh=0;

  int r[4] = {-1,1,0,0};
  int c[4] = {0,0,-1,1};

  queue< pair<int, int>>q;

  for(i=0;i<m;i++) {
    for(j=0;j<n;j++) {
      if (grid[i][j] == 1)
        fresh++;
      else if (grid[i][j] == 2)
        q.push(make_pair(i,j));
    }
  }

  if (fresh == 0)
    return 0;

  if (q.empty())
    return -1;

  for (;;) {
#ifdef DEBUG
    cout << "t = " << min << endl;
    for(i=0;i<m;i++) {
      for(j=0;j<n;j++)
        cout << grid[i][j];
      cout << endl;
    }
    cout << endl;
#endif
    queue< pair<int, int>>qnext;
    while (!q.empty()) {
      pair<int,int> p = q.front();
      int a = p.first;
      int b = p.second;
      q.pop();
      for(i=0;i<4;i++) {
        for(j=0;j<4;j++) {
          int rr = a + r[i];
          int cc = b + c[j];

          if (!(rr<0 || cc<0 || rr>=m || cc>=n || grid[rr][cc]==0 || grid[rr][cc] ==2)
              && (grid[rr][cc] ==1)) {
            grid[rr][cc] = 2;
            qnext.push(make_pair(rr,cc));
            fresh--;  
          }     
        }
      }    
    }
    min += 1;
    if (fresh == 0) {
#ifdef DEBUG
      cout << "t = " << min << endl;
      for(i=0;i<m;i++) {
        for(j=0;j<n;j++)
          cout << grid[i][j];
        cout << endl;
      }
      cout << endl;
#endif   
      return min;
    }
    if (qnext.empty())
      return -1;
    q = qnext;
  } 
}

int main()
{
  vector<vector<int> > grid;

  grid.resize(3);

  grid[0].push_back(2);
  grid[0].push_back(1);
  grid[0].push_back(1);

  grid[1].push_back(1);
  grid[1].push_back(1);
  grid[1].push_back(0);

  grid[2].push_back(0);
  grid[2].push_back(1);
  grid[2].push_back(1);

  cout << orangesRotting(grid) << endl;
}

编译和执行：

/tmp % g++ -DDEBUG oo.cc
/tmp % ./a.out
t = 0
211
110
011

t = 1
221
220
011

t = 2
222
220
022

2

请注意，这种方式比我下面的方式更有效，因为您只考虑每个烂橙子一次

所需时间取决于腐烂橙子周围是否考虑对角线以使新鲜橙子也腐烂。

在我的实现中，我使用预处理器变量DIAG来考虑或不考虑对角线，并使用 DEBUG 每分钟打印或不打印网格：

#include <iostream>
#include <vector>

using namespace std;

enum State { Empty, Fresh, Rotten };

// I do not see the interest of the class so I removed it
// I do not want to modify the input vector so I get it by value

int orangesRotting(vector<vector<State>> grid)
{
  int nmins = 0;
  const size_t height = grid.size();

  if (height == 0)
    return -1;

  const size_t width = grid[0].size(); // suppose same size for all sub vectors

  if (width == 0)
    return -1;

  // the grid for the next min, do not work on the
  // current grid to not see the cells becoming rotten
  // in the current step, changes are done simultaneously
  vector<vector<State>> nextGrid = grid;

  for (;;) {
#ifdef DEBUG
    cout << "t = " << nmins << endl;
#endif

    bool modified = false;
    int nWasFresh = 0;

    for (size_t i = 0; i != height; ++i) {
      vector<State> & v = grid[i];

      for (size_t j = 0; j != width; ++j) {
#ifdef DEBUG
        cout << v[j];
#endif
        switch (v[j]) {
        case Rotten: 
          {
            // make fresh cells around rotten
#ifdef DIAG
            const size_t maxh = min(i + 1, height - 1);
            const size_t minw = (j == 0) ? j : j - 1;
            const size_t maxw = min(j + 1, width - 1);

            for (size_t a = (i == 0) ? i : i - 1; a <= maxh; ++a) {
              vector<State> & v = grid[a];

              for (size_t b = minw; b <= maxw; ++b) {
                if (v[b] == Fresh) {
                  modified = true;
                  nextGrid[a][b] = Rotten;
                }
              }
            }
#else
            if ((i != 0) && (grid[i-1][j] == Fresh)) {
              modified = true;
              nextGrid[i-1][j] = Rotten;
            }
            if ((i != (height-1)) && (grid[i+1][j] == Fresh)) {
              modified = true;
              nextGrid[i+1][j] = Rotten;
            }
            if ((j != 0) && (grid[i][j-1] == Fresh)) {
              modified = true;
              nextGrid[i][j-1] = Rotten;
            }
            if ((j != (width-1)) && (grid[i][j+1] == Fresh)) {
              modified = true;
              nextGrid[i][j+1] = Rotten;
            }
#endif
          }
          break;
        case Fresh:
          nWasFresh += 1;
          break;
        default:
          break;
        }
      }
#ifdef DEBUG
      cout << endl;
#endif
    }
#ifdef DEBUG
    cout << endl;
#endif

    if (nWasFresh == 0)
      return nmins;

    if (!modified)
      return -1;

    // update grid and time
    grid = nextGrid;
    nmins += 1;
  }
}

int main()
{
  vector<vector<State>> grid;

  grid.resize(3);

  grid[0].push_back(Rotten);
  grid[0].push_back(Fresh);
  grid[0].push_back(Fresh);

  grid[1].push_back(Fresh);
  grid[1].push_back(Fresh);
  grid[1].push_back(Empty);

  grid[2].push_back(Empty);
  grid[2].push_back(Fresh);
  grid[2].push_back(Fresh);

  cout << orangesRotting(grid) << endl;
}

考虑对角线的编译和执行：

pi@raspberrypi:/tmp $ g++ -DDEBUG -DDIAG -pedantic -Wextra -Wall o.cc
pi@raspberrypi:/tmp $ ./a.out
t = 0
211
110
011

t = 1
221
220
011

t = 2
222
220
022

2

不考虑对角线的编译和执行：

pi@raspberrypi:/tmp $ g++ -DDEBUG -pedantic -Wextra -Wall o.cc
pi@raspberrypi:/tmp $ ./a.out
t = 0
211
110
011

t = 1
221
210
011

t = 2
222
220
011

t = 3
222
220
021

t = 4
222
220
022

4
于 2019-04-09T17:14:28.047 回答
0

这应该解决它：

public class Orange
{
    public int TimeFrame,x,y;
    public Orange(int x, int y, int timeFrame)
    {
        this.x = x;
        this.y = y;
        this.TimeFrame = timeFrame;

    }
    
}
internal class RottenOrange
{
    Queue<Orange> queue = new Queue<Orange>();
    int timeFrame = 0;

    public int OrangesRotting(int[][] grid)
    {
        
        FindRottenOrange(grid,true);
        while(queue.Count>0)
        {
            var dequeue = queue.Dequeue();
            // check left
            if( dequeue.y>0)
            {
                if (grid[dequeue.x][dequeue.y - 1] == 1)
                {
                    grid[dequeue.x][dequeue.y - 1] = 2;
                    queue.Enqueue(new Orange(dequeue.x, dequeue.y - 1, dequeue.TimeFrame + 1));
                    timeFrame = dequeue.TimeFrame + 1;
                }
            }
            // check right
            if (dequeue.y < grid[dequeue.x].Length-1)
            {
                if (grid[dequeue.x][dequeue.y+1] == 1)
                {
                    grid[dequeue.x][dequeue.y + 1] = 2;
                       queue.Enqueue(new Orange(dequeue.x, dequeue.y+1, dequeue.TimeFrame + 1));
                    timeFrame = dequeue.TimeFrame + 1;
                }
                

            }
            // check up
            if (dequeue.x >0)
            {
                if (grid[dequeue.x - 1][dequeue.y] == 1)
                {
                    grid[dequeue.x - 1][dequeue.y] = 2;
                    queue.Enqueue(new Orange(dequeue.x - 1, dequeue.y, dequeue.TimeFrame + 1));
                    timeFrame = dequeue.TimeFrame + 1;
                }

            }
            // check down
            if (dequeue.x < grid.Length-1)
            {
                if (grid[dequeue.x+1][dequeue.y] == 1)
                {
                    grid[dequeue.x + 1][dequeue.y] = 2;
                    queue.Enqueue(new Orange(dequeue.x+1, dequeue.y, dequeue.TimeFrame + 1));
                    timeFrame = dequeue.TimeFrame + 1;
                }

            }

        }
        FindRottenOrange(grid,false);
        if(queue.Count>0)
        {
            timeFrame = -1;
        }
        return timeFrame;
    }
    private Queue<Orange> FindRottenOrange(int[][] grid,bool isRotten)
    {

        for (int i = 0; i < grid.Length; i++)
        {
            for (int j = 0; j < grid[i].Length; j++)
            {
                if (isRotten && grid[i][j] == 2)
                {
                    queue.Enqueue(new Orange(i, j, timeFrame));
                }
                else if(!isRotten && grid[i][j] == 1)
                {
                    queue.Enqueue(new Orange(i, j, timeFrame));
                }
            }
        }

        return queue;


    }
}
于 2021-12-10T19:49:18.420 回答