Leetcode463 Island Perimeter

very easy to understand:
Input:
[[0,1,0,0],
[1,1,1,0],
[0,1,0,0],
[1,1,0,0]]

Output: 16

the modified version of flood fill alg, but we need to transfer out mind a little bit.

first, we know that this problem needs to be solved in dfs.
then think of a lonely 1*1 island.
starting from this island, if left of it is water/out of bound, then the left edge belongs to us, count+1
moving to right, if right part is island as well, then this right edge can’t be count as an edge, so count+0
and only when this direction is an island will we continue our dfs.

public class Solution {
    public int islandPerimeter(int[][] grid) {
        if (grid == null ||grid[0] == null || grid.length == 0) {
            return 0;
        }
        
        int m = grid.length;
        int n = grid[0].length;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    return perimeter(grid, i, j);
                }
            }
        }
        return 0;
    }
    
    private int perimeter(int[][] grid, int i, int j) {
        int count = 0; //top level
        
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length) {
            return 1;
        }      
        if(grid[i][j] == 0) {
            return 1;
        }
        if (grid[i][j] == -1) { //means we revisied it, and this line will not exist
            return 0;
        }
        
        grid[i][j] = -1; //mark here to show here is a land, and we don't have edge
        count += perimeter(grid, i - 1, j);
        count += perimeter(grid, i + 1, j);
        count += perimeter(grid, i, j - 1);
        count += perimeter(grid, i, j + 1);
        return count;
    }
}