Number of Islands

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

/*

When find one land, increment the number of lands, then use DFS/BFS to find all the adjacent lands. In order to avoid duplicate, set each land just found to be water, like set '1' to '0'.

*/

//DFS, time:O(m*n), space:O(1)

public class Solution {

    public int numIslands(char[][] grid) {

        if (grid == null || grid.length == 0 || 

            grid[0] == null || grid[0].length == 0) {

            return 0;

        }

        int m = grid.length;

        int n = grid[0].length;

        int count = 0;

        for (int i = 0; i < m; i++) {

            for (int j = 0; j < n; j++) {

                if (grid[i][j] == '0') {

                    continue;

                }

                count++;

                dfs(grid, i, j);

            }

        }

        return count;

    }

    public void dfs(char[][] grid, int i, int j) {

        if (i < 0 || i >= grid.length || 

            j < 0 || j >= grid[0].length) {

            return;

        }

        if (grid[i][j] == '1') {

            grid[i][j] = '0';

            dfs(grid, i + 1, j);

            dfs(grid, i, j + 1);

            dfs(grid, i - 1, j);

            dfs(grid, i, j - 1);

        }

    }

}

//BFS, time:O(m*n), space:O(m*n)

public class Solution {

    public int numIslands(char[][] grid) {

        if (grid == null || grid.length == 0 || 

            grid[0] == null || grid[0].length == 0) {

            return 0;

        }

        int m = grid.length;

        int n = grid[0].length;

        int count = 0;

        for (int i = 0; i < m; i++) {

            for (int j = 0; j < n; j++) {

                if (grid[i][j] == '0') {

                    continue;

                }

                count++;

                bfs(grid, i, j);

            }

        }

        return count;

    }

    public void bfs(char[][] grid, int i, int j) {

        int m = grid.length;

        int n = grid[0].length;

        int[] dx = {1, 0, -1, 0};

        int[] dy = {0, 1, 0, -1};

        Queue<Integer> queue = new LinkedList<>();

        queue.offer(i * n + j);

        grid[i][j] = '0';

        while (!queue.isEmpty()) {

            int index = queue.poll();

            for (int k = 0; k < 4; k++) {

                int nx = index / n + dx[k];

                int ny = index % n + dy[k];

                if (nx < 0 || nx >= m || ny < 0 || ny >= n) {

                    continue;

                }

                if (grid[nx][ny] == '0') {

                    continue;

                }

                queue.offer(nx * n + ny);

                grid[nx][ny] = '0';

            }

        }

    }

}