[leetcode-695-Max Area of Island]

最新推荐文章于 2022-02-26 13:10:39 发布

转载最新推荐文章于 2022-02-26 13:10:39 发布 · 83 阅读

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原文链接：http://www.cnblogs.com/hellowooorld/p/7636814.html

本文介绍了一种使用深度优先搜索(DFS)算法来解决二维数组中寻找最大岛屿面积的问题。通过遍历每个元素并利用递归的方式标记已访问过的陆地部分，最终找到最大的连续1（代表陆地）组成的岛屿面积。

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Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.

思路：

DFS即可。

 int a[] = {-1,0,0,1};
 int b[] = {0,1,-1,0};
 void DFS(int m,int n,int curM,int curN,vector<vector<int>>& grid,int &temp,int &ret,vector<vector<bool>>&visited)
 {
   if(curM<0 || curN<0 || curM>=m || curN>=n || visited[curM][curN] || grid[curM][curN] ==0 )return ;
   temp+=1;
   visited[curM][curN] = true;
   for(int i=0;i<4;i++)
   {
     DFS(m,n,curM+a[i],curN+b[i],grid,temp,ret,visited);
  }
   ret = max(temp,ret);   
} 
 int maxAreaOfIsland(vector<vector<int>>& grid)
 {
        int m =grid.size(),n=0;
    if(m!=0)n = grid[0].size();
    vector<vector<bool>>visited(m,vector<bool>(n,false));
    int temp =0,ret =0;
    for(int i=0;i<m;i++)
    {
      for(int j =0;j<n;j++)
      {
        DFS(m,n,i,j,grid,temp,ret,visited);
        temp  = 0;
      }
    }    
    return ret;
 }

转载于:https://www.cnblogs.com/hellowooorld/p/7636814.html