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
Example 2:

11000
11000
00100
00011

Answer: 3

Naive Way: Use DFS to go though all grid cells. Use a hashset to keep visited grid cells. Run time is O(n^2). And space is O(n^2).

 public class Solution {  
   public int numIslands(char[][] grid) {  
     Set<List<Integer>> visited = new HashSet<List<Integer>>();  
     int num = 0;  
     for(int i = 0;i < grid.length;i++)  
       for(int j = 0;j < grid[0].length;j++)  
         num+=dfs(grid,i,j,visited);  
     return num;  
   }  
   private int dfs(char[][] grid, int x, int y, Set<List<Integer>> visited){  
     // edge case  
     if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) return 0;  
     // island  
     if(grid[x][y] == '1'){  
       List<Integer> list = new ArrayList<Integer>();  
       list.add(x);  
       list.add(y);  
       if(!visited.contains(list)){  
         visited.add(list);  
         dfs(grid, x-1, y, visited);  
         dfs(grid, x+1, y, visited);  
         dfs(grid, x, y-1, visited);  
         dfs(grid, x, y+1, visited);  
         return 1;  
       }  
     }  
     return 0;  
   }  
 }  

I saw a algorithm in Discuss use the original grid to store visited grid cells. Thus reduce space used to O(1).