695. Max Area of Island

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.

题意要求，给定一个二维数组，从四个方向上链接的（用1来表示）的可以算做一个岛屿，求面积最大的岛屿的面积。

这个题就是用DFS，可以用BFS试一下，在这里立一个flag!

class Solution {public:    int maxAreaOfIsland(vector<vector<int>>& grid) {        if (grid.size() == 0 || grid[0].size() == 0) return 0;        int maxArea = 0;        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), false));        for (int i = 0; i < grid.size(); i++) {            for (int j = 0; j < grid[i].size(); j++) {                if (grid[i][j] == 1 || visited[i][j] == false) {                    int result = 0;                    DFS(grid, visited, i, j, result);                    maxArea = max(maxArea, result);                }            }        }        return maxArea;       //  [1,1,0,0,0]       //  [1,1,0,0,0]       //  [0,0,0,1,1]       //  [0,0,0,1,1]    }    void DFS(vector<vector<int>>& grid, vector<vector<bool>>& visited, int i, int j, int& result) {        if (i < 0 || j < 0 || i >= grid.size() || j >= grid[i].size() || grid[i][j] == 0 || visited[i][j] == true) {            return;        } else {            if (visited[i][j] == false) {                visited[i][j] = true;                result++;                DFS(grid, visited, i + 1, j    , result);//down                DFS(grid, visited, i    , j + 1, result);//right                DFS(grid, visited, i - 1, j    , result);//up                DFS(grid, visited, i    , j - 1, result);//left            }        }    }};