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 <= grid.length, grid[0].length <= 100​​
• ​​0 <= grid[i][j] <=1​​

​​题解：​​

class Solution {
public:
    int next[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    void dfs(int x, int y, int m, int n, vector<vector<int>> & grid, bool &island) {
        if (x == 0 || x == m - 1 || y == 0 || y == n - 1) {
            island = false;
        }
        for (int i = 0; i < 4; i++) {
            int nx = x + next[i][0], ny = y + next[i][1];
            if (nx >= 0 && nx < m && ny >= 0 && ny < n) {
                if (grid[nx][ny] == 0) {
                    grid[nx][ny] = 1;
                    dfs(nx, ny, m, n, grid, island);
                }
            }
        }
    }
    int closedIsland(vector<vector<int>>& grid) {
        if (grid.empty() == true) {
            return 0;
        }
        int m = grid.size(), n = grid[0].size();
        bool island = true;
        int res = 0;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) {
                    grid[i][j] = 1;
                    island = true;
                    dfs(i, j, m, n, grid, island);
                    if (island == true) {
                        res++;
                    }
                }
            }
        }
        return res;
    }
};