LeetCode： Max Area of Island

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：

[[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.

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

Given the above grid, return 0.

思路 ：

这是一道DFS深度优先的题。
直接对grid中各个值为1的格子进行深度优先搜索，求出邻接的面积，然后取最大值。

代码：

class Solution {public:    int maxAreaOfIsland(vector<vector<int>>& grid) {        int row = grid.size();        int col = grid[0].size();        int sum = 0;        for (int i = 0; i < row; i++) {            for (int j = 0; j < col; j++) {                if (grid[i][j] == 1) {                    sum = max(sum, dfs(grid, i, j));                }            }        }        return sum;    }private:    int dfs(vector<vector<int>>& grid, int row, int col) {        if (row < 0 || col < 0 || row >= grid.size() || col >= grid[0].size() || grid[row][col] != 1) {            return 0;        }        int area = 1;        grid[row][col] = 2;        area += dfs(grid, row, col+1);        area += dfs(grid, row+1, col);        area += dfs(grid, row, col-1);        area += dfs(grid, row-1, col);        return area;    }};