200.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:

Input:

11110
11010
11000
00000


Output:
 1

Example 2:

Input:

11000
11000
00100
00011


Output: 
3

2D array direction problem

grid[i][j] i is y axis, j is x axis

Concise solution

class Solution {
    public void dfs(char[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i == grid.length || j == grid[i].length || grid[i][j] != '1') return;
        grid[i][j] = '0';
        dfs(grid, i, j+1);
        dfs(grid, i+1, j);
        dfs(grid, i, j-1);
        dfs(grid, i-1, j);
    }

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j);
                    count++;
                }
            }
        }
        return count;
    }
}

My solution

class Solution {
    public void dfs(char[][] grid, int y, int x) {
        grid[y][x] = 'V';
        // East
        if (x+1 < grid[y].length && grid[y][x+1] == '1')
            dfs(grid, y, x+1);
        // South
        if (y+1 < grid.length && grid[y+1][x] == '1')
            dfs(grid, y+1, x);
        // West
        if (x-1 >= 0 && grid[y][x-1] == '1')
            dfs(grid, y, x-1);
        // North
        if (y-1 >= 0 && grid[y-1][x] == '1')
            dfs(grid, y-1, x);
    }

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j);
                    count++;
                }
            }
        }
        return count;
    }
}