Given a 2Dgridconsists of0s(land)and1s(water). Anislandis a maximal 4-directionally connected group of0sand aclosed islandis an islandtotally(all left, top, right, bottom) surrounded by1s.

Return the number ofclosed 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 <= grid.length, grid[0].length <= 100
• 0 <= grid[i][j] <=1

解法：其中0的部分為island，但是接觸到四個邊界的island不算，只要按照這個規則進行dfs即可

class Solution {
    public int closedIsland(int[][] grid) {
        int count = 0;
        for( int i=0;i<grid.length;i++){
            for(int j=0;j<grid[0].length;j++){
                
 
if(grid[i][j]==0 && dfs(grid,i,j)) {
                    count++;
                }
            }
        }
        return count;
    }
    private boolean dfs(int[][] grid,int i,int j){
        if(i<0||j<0||i>=grid.length||j>=grid[0].length) return false;
        if(grid[i][j]!=0) return true;
        grid[i][j]=1;
        return dfs( grid,i-1,j) & dfs(grid,i+1,j) & dfs(grid,i,j-1) & dfs(grid,i,j+1);
    }
}