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.

 

dfs

设置一个global variable area表示岛的面积。遍历数组元素，如果遇到1，把area重置为0，dfs遍历其上下左右求面积，再与max比较取较大值作为结果。

dfs：终止条件是数组下标越界或者该点不为1。如果该点值为1，面积area增加1，把当前点赋值为0表示已访问过，然后遍历dirs数组对其上下左右的点进行同样的dfs操作。

时间：O(M*N)，空间：O(M*N) for worst case dfs recursion depth

class Solution {    int[][] dirs = {
   {-1, 0}, {0, 1}, {1, 0}, {0, -1}};  //top, right, down, left    int area;        public int maxAreaOfIsland(int[][] grid) {        if(grid == null || grid.length == 0) return 0;        int max = 0;        for(int i = 0; i < grid.length; i++) {            for(int j = 0; j < grid[0].length; j++) {                if(grid[i][j] == 1) {                    area = 0;                    dfs(grid, i, j);                    max = Math.max(max, area);                }            }        }        return max;    }        private void dfs(int[][] grid, int x, int y) {        if(x < 0 || x > grid.length - 1 || y < 0 || y > grid[0].length - 1 || grid[x][y] == 0) return;        area += 1;        grid[x][y] = 0;        for(int[] dir : dirs) {            int xpos = x + dir[0];            int ypos = y + dir[1];            dfs(grid, xpos, ypos);        }    }}