leetcode

first version

First version is straight forward, check each unit, and check the four units around it. If they are valid island, continue. else, perimeter increment.

public class Solution {
    int[][] grid;
    int row;
    int col;

    public int islandPerimeter(int[][] grid) {
        this.grid = grid;
        row = grid.length;
        if(row == 0)return 0;
        col = grid[0].length;

        int perimeter = 0;
        for(int i = 0; i < row; i ++){
            for(int j = 0; j < col; j++){
                if(!isIsland(i,j))continue;
                if(!isIsland(i - 1, j))perimeter++;
                if(!isIsland(i + 1, j))perimeter++;
                if(!isIsland(i, j - 1))perimeter++;
                if(!isIsland(i, j + 1))perimeter++;
            }
        }
        return perimeter;

    }

    public boolean isIsland(int i, int j){
        if(i < 0 || i >= row || j < 0 || j >= col)return false;
        return grid[i][j] == 1;
    }


}

Other versions

I see some top solutions on leetcode discussion, and the idea is the same. some optimization is to count 4 for each unit, and when it has a neighbors( right and down directions to avoid repeat), then minus 2, since the two neighbors will share an edge.

see this solution