695. Max Area of Island

Given a non-empty 2D array grid of 0‘s and 1‘s, an island is a group of 1‘s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

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

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

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

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.

//Time: O(m * n), Space: O(m * n)
//如果可以改变原数组，可以不用visited，每次遍历完1的位置改成0即可，这样就不用开辟额外空间
public int maxAreaOfIsland(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        
        boolean[][] visited = new boolean[grid.length][grid[0].length];
        int area = 0;
        
        for (int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                     area = Math.max(dfs(i, j, grid, visited), area);
                }
            }
        }
        
        return area;
    }
    
    private int dfs(int i, int j, int[][] grid, boolean[][] visited){
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || visited[i][j] == true || grid[i][j] == 0) {
            return 0;
        }
        
        visited[i][j] = true;
        return 1 + dfs(i + 1, j, grid,visited) + dfs(i - 1, j, grid, visited) + dfs(i, j + 1, grid, visited) + dfs(i, j - 1, grid, visited);
    }