695. Max Area of Island最大岛屿面积

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]]

思路：加一个1就消灭一个，记为0，各个击破。消灭了就消灭了，联通的一个岛统计一次也就够了

要限制范围在数组的length以内，如果不符合条件就返回0

还要逐个比较max = max(max, value)得出最大值

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        //corner case
        if (grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        //compare all areas
        int max = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                max = Math.max(max, areaOfIsland(i, j, grid));
            }
        }
        return max;
    }
    
    public int areaOfIsland(int i, int j, int[][] grid) {
        //valid first, == 1 second
    if (0 <= i && i < grid.length && 0<= j && j < grid[0].length && grid[i][j] == 1) {
        //restore to 0 to avoid repeat
        grid[i][j] = 0;
        //count area
        return 1 + areaOfIsland(i - 1, j, grid) + areaOfIsland(i + 1, j, grid) + areaOfIsland(i, j - 1, grid) + areaOfIsland(i, j + 1, grid);
    }
        //if not 1, default case : return 0
        return 0;
    }
}