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 <= grid.length, grid[0].length <= 100

• 0 <= grid[i][j] <=1

Solution

先把四個邊都DFS過。

 public int closedIsland(int[][] grid) {
        int M = grid.length;
        int N = grid[0].length;
        for(int i = 0; i < M; i++){
            dfs(grid, i, 0);
            dfs(grid, i, N-1);
        }
        
        for(int j = 0; j < N; j++){
            dfs(grid, 0, j);
            dfs(grid, M-1, j);
        }
        int count = 0;
        for(int i = 1; i < M; i++){
            for(int j = 1; j < N; j++){
                if(grid[i][j] == 0){
                    dfs(grid, i, j);
                    count++;
                }
            }
        }
        return count;
    }
    
    public void dfs(int[][] grid, int i, int j){
        int M = grid.length;
        int N = grid[0].length;
        
        if(i < 0 || i >= M || j < 0 || j >= N || grid[i][j] == 1) return;
        grid[i][j] = 1;
        int[][] dirs = new int[][]{{0,1}, {1,0}, {0,-1}, {-1,0}};
        
        for(int[] dir : dirs){
            int x = i + dir[0];
            int y = j + dir[1];
            dfs(grid, x, y);
        }
        
    }