Leetcode_1254_Number of Closed Islands

[lihe]

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:
    void dfs(std::vector<std::vector<int>> &grid, int i, int j, int &val){
        if(i < 0 || i == grid.size()|| j < 0 || j == grid[0].size()){
            val = 0;
            return;
        }
        if(grid[i][j] == 1)
            return;
        grid[i][j] = 1;
        dfs(grid, i + 1, j, val);
        dfs(grid, i - 1, j, val);
        dfs(grid, i, j - 1, val);
        dfs(grid, i, j + 1, val);
    }

    int closedIsland(std::vector<std::vector<int>> &grid){
        int result = 0;
        for(int i = 0; i < grid.size(); i++){
            for(int j = 0; j < grid[i].size(); j++){
                if(grid[i][j] == 0){
                    int val = 1;
                    dfs(grid, i, j, val);
                    result += val;
                }
            }
        }
        return result;
    }
};