You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn’t have “lakes” (water inside that isn’t connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don’t exceed 100. Determine the perimeter of the island.

Example:

[[0,1,0,0],
 [1,1,1,0],
 [0,1,0,0],
 [1,1,0,0]]

Answer: 16
Explanation: The perimeter is the 16 yellow stripes in the image below:
public class Solution {
    
    private int result = 0;
    private final int[] dx = new int[]{-1,0,1,0};
    private final int[] dy = new int[]{0,1,0,-1};
    private boolean[][] visited;
    
    public int islandPerimeter(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
        visited = new boolean[grid.length][grid[0].length];
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    bfs(i, j, grid);
                }
                
            }
        }
        return result;
    }
    
    private void bfs(int x, int y, int[][] grid) {
        visited[x][y] = true;
        Queue<int[]> q = new LinkedList<>();
        q.offer(new int[]{x, y});
        while (!q.isEmpty()) {
            int[] pos = q.poll();
            for (int i = 0; i < 4; i++) {
                int nx = pos[0] + dx[i];
                int ny = pos[1] + dy[i];
                if (nx >= 0 && nx < grid.length && ny >= 0 && ny < grid[0].length) {
                    if (grid[nx][ny] == 1 && !visited[nx][ny]) {
                        q.offer(new int[]{nx, ny});
                        visited[nx][ny] = true;
                    }
                    else if (grid[nx][ny] == 0) result++;
                } else {
                    result++;
                }
            }
        }
    }
}

// DFS
public class Solution {
    
    private int result = 0;
    private final int[] dx = new int[]{-1,0,1,0};
    private final int[] dy = new int[]{0,1,0,-1};
    private boolean[][] visited;
    
    public int islandPerimeter(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
        visited = new boolean[grid.length][grid[0].length];
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    dfs(i, j, grid);
                    return result;
                }
                
            }
        }
        return result;
    }
    
    private void dfs(int x, int y, int[][] grid) {
        visited[x][y] = true;
        for (int i = 0; i < 4; i++) {
            int nx = x + dx[i];
            int ny = y + dy[i];
            if (nx >= 0 && nx < grid.length && ny >= 0 && ny < grid[0].length) {
                if (grid[nx][ny] == 1 && !visited[nx][ny]) {
                    visited[nx][ny] = true;
                    dfs(nx, ny, grid);
                } else if (grid[nx][ny] == 0) result++;
            } else result++;
        }
    }
}
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