1254. Number of Closed Islands

JIAKAOBO

LeetCode

感谢赞助!

• ㊗️
• 大家
• offer
• 多多！

Problem

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

Code

class Solution {
    int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    public int closedIsland(int[][] grid) {
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(i == 0 || j == 0 || i == grid.length - 1 || j == grid[i].length - 1) {
                    dfs(grid, i, j);
                }
            }
        }

        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) {
                    dfs(grid, i, j);
                    res++;
                }
            }
        }

        return res;
    }

    private void dfs(int[][] grid, int x, int y) {
        if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == 1) return;

        grid[x][y] = 1;

        for(int[] dir : dirs) {
            dfs(grid, x + dir[0], y + dir[1]);
        }
    }
}