LeetCode 695. Max Area of Island

最新推荐文章于 2024-07-29 23:53:47 发布

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本文介绍了一个算法问题：在一个由0和1组成的二维数组中寻找最大的岛屿面积，岛屿由相邻的1（代表陆地）组成。文章详细解释了如何通过深度优先搜索(DFS)遍历每个可能的岛屿并计算其面积，最终返回所有岛屿中的最大面积。

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.

int maxAreaOfIsland(vector<vector<int>>& grid) {
int m=grid.size();
int n=grid[0].size();
vector<vector<int>> ischeck(m,vector<int>(n,0));
int res=0;
for(int i=0;i<m;++i)
{
for(int j=0;j<n;++j)
{
res=max(res,dfs(grid,ischeck,i,j));
}
}
return res;
}

int dfs(vector<vector<int>>& grid ,vector<vector<int>>& ischeck ,int i,int j)
{
int num=0;
if(i<0||i>=grid.size()||j<0||j>=grid[0].size())
return 0;
if(ischeck[i][j]==0&& grid[i][j]==1)
{
ischeck[i][j]=1;
num+=1;
num+=dfs(grid,ischeck,i+1,j);
num+=dfs(grid,ischeck,i-1,j);
num+=dfs(grid,ischeck,i,j+1);
num+=dfs(grid,ischeck,i,j-1);
}
return num;

}