题目：

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.

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题目的意思就是：在一个二维数组中找到相邻1的个数，返回最大的个数

思路：这个问题有点像五子棋中判断五子的方法，通过递归就很好实现。注意一点就是：为了不重复访问相同区域，只要访问过的点就可以变为0。

直接贴上代码：

class Solution {private:　　int Maxnum(vector
     
      
       >& temp, int i, int j)　　{　　　　//递归的条件　　　　if ( i >= 0 && i < temp.size() && j >= 0 && j < temp[0].size() &&           /*注意这个条件放最后，放最前面会越界访问*/temp[i][j] == 1)　　　　{　　　　　　temp[i][j] = 0;　　　　//访问过的重置为0　　　　　　return Maxnum(temp, i + 1, j) + Maxnum(temp, i - 1, j) +                  Maxnum(temp, i, j + 1) + Maxnum(temp, i, j - 1) + 1;　　//递归　　　　}　　　　return 0;　　}public:　　int maxAreaOfIsland(vector
       
        
         >& grid) 　　{　　　　int max = 0;　　　　for (int i = 0; i < grid.size(); ++i)　　　　{　　　　　　　for (int j = 0; j < grid[0].size(); ++j)　　　　　　{　　　　　　　　if (grid[i][j] == 1)　　　　　　　　　　max = std::max(Maxnum(grid, i, j), max);    //std::max是标准库中max函数　　　　　　}　　　　}　　　　return max;　　}};