Number of Island -- Leetcode

Question:

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

Example 2:

11000
11000
00100
00011

Answer: 3

Answer:
1. Inplace, Time: O(m*n), Space:O(1)

public class Solution {

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

        if(grid==null)return 0;

        int m = grid.length;

        if(m==0)return 0;

        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]=='1'){

                    count++;

                    DFS(grid, i,j);

                }

            }

        }

        return count;

    }

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

        int m=grid.length, n=grid[0].length;

        if(i<0||i>=m||j<0||j>=n)return;

        if(grid[i][j]!='1'){

            return;

        }

        grid[i][j]='2';

        DFS(grid,i-1,j);

        DFS(grid,i,j-1);

        DFS(grid,i+1,j);

        DFS(grid,i,j+1);

        return;

    }

}


2.  Time: O(m*n), Space:O(m*n), using additional boolean[m][n] flag.
public class Solution {
    public int numIslands(char[][] grid) {
        if(grid==null)return 0;
        int m = grid.length;
        if(m==0)return 0;
        int n = grid[0].length;
        int count=0;
        boolean[][] flag;
       
        flag = new boolean[m][];
        for(int i=0;i<m;++i){
            flag[i] = new boolean[n];
            for(int j=0;j<n;++j){
                flag[i][j]=false;
            }
        }
               
        for(int i=0;i<m;++i){
            for(int j=0;j<n;++j){
                if(grid[i][j]=='1' && flag[i][j]==false){
                    count++;
                    DFS(grid,i,j,flag);
                }
            }
        }
        return count;
    }

    public void DFS(char[][] grid, int i, int j, boolean[][] flag){
        //check valid boundary
        int m=grid.length, n=grid[0].length;
        if(i<0||i>=m||j<0||j>=n)return;
       
        if(grid[i][j]=='1' && flag[i][j]==false){
            flag[i][j]=true;
            DFS(grid,i-1,j,flag);
            DFS(grid,i,j-1,flag);
            DFS(grid,i+1,j,flag);
            DFS(grid,i,j+1,flag);
        }
        return;
    }
}