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

class Solution {
    public int closedIsland(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;        
        int res = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 0 && i == 0 || j == 0 || i == m - 1 || j == n - 1) help(i, j, grid);
            }
        }
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 0){
                  res++;
                  help(i, j, grid);  
                }
            }
        }
        return res;
    }
    public void help(int r, int c, int[][] grid){
        if(r > -1 && r < grid.length && c > -1 && c < grid[0].length && grid[r][c] == 0) grid[r][c] = 1;
        else return;
        help(r + 1, c, grid);
        help(r, c + 1, grid);
        help(r - 1, c, grid);
        help(r, c - 1, grid);
    }
}

DFS, 先把边界的0区域全置1，然后再遍历，遇到0就先加1，然后再把该0区域置1，最后得到答案。

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

如果要把所有周围的岛涂成1，一定要一个一个来，如果用&& ||，尤其是||，可能会出问题，因为有可能涂到一半就不涂了，四个方向one by one可以避免这个问题。

或者

class Solution {

    int[][] dir = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    
    public int closedIsland(int[][] grid) {
        int res = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 0){
                    if(dfs(grid, i, j)) res++;
                }
            }
        }
        
        return res;
    }
    
    public boolean dfs(int[][] grid, int x, int y){
        
        if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) return false;
        
        if(grid[x][y] == 1) return true;
        
        grid[x][y] = 1;
        
        return dfs(grid, x + 1, y) & dfs(grid, x, y + 1) & dfs(grid, x - 1, y) & dfs(grid, x, y - 1);
    }
}