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  • Leetcode-994 Rotting Oranges(腐烂的橘子)

     1 #define _for(i,a,b) for(int i = (a);i < (b);i ++)
     2 class Solution
     3 {
     4     public:
     5         int orangesRotting(vector<vector<int>>& grid)
     6         {
     7             int left = 0;
     8             int sz1 = grid.size();
     9             int sz2 = grid[0].size();
    10             _for(i,0,sz1)
    11                 _for(j,0,sz2)
    12                     if(grid[i][j]==1)
    13                         left ++;
    14             
    15             int prev = left;
    16             int rnt = 0;
    17             while(left)
    18             {
    19                 prev = left;
    20                 _for(i,0,sz1)
    21                     _for(j,0,sz2)
    22                     {
    23                         if(grid[i][j]==2)
    24                         {
    25                             if(i+1<sz1&&grid[i+1][j]==1)
    26                             {
    27                                 left--;
    28                                 grid[i+1][j]=-1;
    29                             }
    30                             if(i-1>=0&&grid[i-1][j]==1)
    31                             {
    32                                 left--;
    33                                 grid[i-1][j]=-1;
    34                             }
    35                             if(j-1>=0&&grid[i][j-1]==1)
    36                             {
    37                                 left--;
    38                                 grid[i][j-1]=-1;
    39                             }
    40                             if(j+1<sz2&&grid[i][j+1]==1)
    41                             {
    42                                 left--;
    43                                 grid[i][j+1]=-1;
    44                             }
    45                         }
    46                     }
    47                     
    48                 _for(i,0,sz1)
    49                     _for(j,0,sz2)
    50                         if(grid[i][j]==-1)
    51                             grid[i][j] = 2;
    52                 rnt ++;
    53                 if(prev==left)
    54                     return -1;
    55             }
    56             return rnt;
    57         }
    58 };

     原则上来说,应该用BFS,看了一下数据范围,决定无脑暴力撸过

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  • 原文地址:https://www.cnblogs.com/Asurudo/p/10390642.html
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