Leetcode 1254. Number of Closed Islands

Given a 2D grid consists of 0s (land) and 1s (water). An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s.

Return the number of closed islands.

Example 1:

Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
Output: 2
Explanation:
Islands in gray are closed because they are completely surrounded by water (group of 1s).
Example 2:

Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
Output: 1
Example 3:

Input: grid = [[1,1,1,1,1,1,1],
               [1,0,0,0,0,0,1],
               [1,0,1,1,1,0,1],
               [1,0,1,0,1,0,1],
               [1,0,1,1,1,0,1],
               [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
Output: 2
 

Constraints:

1 
solution
需要找出四周环水的岛屿数量，在边界处的岛屿不能算四周环水，所以先把四周的岛屿变成水，在按照Leetcode 200. Number of Islands的方法处理就好了
class Solution {
public:
    int closedIsland(vector<vector<int>>& grid) {
        if (grid.empty())
			return 0;
		int n = grid.size();
		int m = grid[0].size();
        
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                if(grid[i][j] == 0 && (i == 0 || j == 0 || i == (n-1) || j == (m-1)))
                    DFS(grid, i, j, n, m);
        
		int ans = 0;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
			{
				if (grid[i][j] == 0)
				{
					ans++;
					DFS(grid, i, j, n, m);
				}
			}
		return ans;

	}
	void DFS(vector<vector<int>>& grid, int i, int j, int n, int m)
	{
		if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] == 1)
			return;
		grid[i][j] = 1;
		DFS(grid, i + 1, j, n, m);
		DFS(grid, i - 1, j, n, m);
		DFS(grid, i, j + 1, n, m);
		DFS(grid, i, j - 1, n, m);

	}
    
};

参考链接

leetcode
Leetcode 200. Number of Islands