Perform dfs and add 1 to the perimieter at each terminal condition of dfs. Skip the sharing edges to avoid double counting.

```
class Solution {
public int islandPerimeter(int[][] grid) {
int n = grid.length;
if (n == 0) return 0;
int m = grid[0].length;
int perimeter = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) {
perimeter += dfs(grid, i, j, n, m);
}
}
}
return perimeter;
}
private int dfs(int[][] grid, int i, int j, int n, int m) {
int perimeter = 0;
if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] == 0) {
return 1;
}
if (grid[i][j] == -1) {
return 0;
}
grid[i][j] = -1;
perimeter += dfs(grid, i-1, j, n, m);
perimeter += dfs(grid, i+1, j, n, m);
perimeter += dfs(grid, i, j-1, n, m);
perimeter += dfs(grid, i, j+1, n, m);
return perimeter;
}
}
```