Problem

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.

思路一（divide & conquer）

    public int maxAreaOfIsland(int[][] grid) {
        int max_area = 0;
        for(int i = 0; i < grid.length; i++)
            for(int j = 0; j < grid[0].length; j++)
                if(grid[i][j] == 1)max_area = Math.max(max_area, AreaOfIsland(grid, i, j));
        return max_area;
    }

    public int AreaOfIsland(int[][] grid, int i, int j){
        if( i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
            grid[i][j] = 0;
            return 1 + AreaOfIsland(grid, i+1, j) + AreaOfIsland(grid, i-1, j) + AreaOfIsland(grid, i, j-1) + AreaOfIsland(grid, i, j+1);
        }
        return 0;
    }

思路二（divide & conquer + traverse）

• 带着sum往下， 每次更新sum，到底返回sum
• 因为带着sum， 所以子dfs和父dfs返回的是一样的结果， 其实不是的

    public int maxAreaOfIsland(int[][] grid) {
        if (grid == null || grid.length == 0) return 0;

        int res = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    // dfs (grid, i, j);
                    res = Math.max(res, dfs(grid, i, j, 0));
                }
            }
        }
        return res;
    }

    private int dfs (int[][] grid, int x, int y, int sum) {
        sum++;
        grid[x][y] = 0;
        for (int i = 0; i < 4; i++) {
            int r = x + dx[i];
            int c = y + dy[i];
            if (r < 0 || r >= grid.length || c < 0 || c >= grid[0].length || grid[r][c] == 0) continue;
            sum = dfs(grid, r, c, sum);
        }
        return sum;
    }

    int[] dx = {1, -1, 0, 0};
    int[] dy = {0, 0, 1, -1};