给定一个包含了一些 0 和 1的非空二维数组 grid , 一个 岛屿 是由四个方向 (水平或垂直) 的 1 (代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。

找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为0。)

示例 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]]

对于上面这个给定矩阵应返回 6。注意答案不应该是11，因为岛屿只能包含水平或垂直的四个方向的‘1’。

示例 2:

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

对于上面这个给定的矩阵, 返回 0。

注意: 给定的矩阵grid 的长度和宽度都不超过 50。

BFS：

class Solution {
public:vector<vector<int> > visit;int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int r;int c;int maxAreaOfIsland(vector<vector<int> >& grid) {r = grid.size();if(r == 0)return 0;c = grid[0].size();visit = vector<vector<int> >(r, vector<int>(c, 0));int res = 0;for(int i = 0; i < r; i++){for(int j = 0; j < c; j++){if(visit[i][j] != 1 && grid[i][j] == 1){visit[i][j] = 1;res = max(res, BFS(grid, i, j));}}}return res;}int BFS(vector<vector<int> >& grid, int x, int y){queue<pair<int, int> > q;q.push(make_pair(x, y));int cnt = 0;while(!q.empty()){int xx = q.front().first;int yy = q.front().second;cnt++;q.pop();for(int i = 0; i < 4; i++){int newx = xx + dx[i];int newy = yy + dy[i];if(newx < 0 || newx >= r || newy < 0 || newy >= c)continue;if(visit[newx][newy] == 1)continue;if(grid[newx][newy] == 0)continue;visit[newx][newy] = 1;q.push(make_pair(newx, newy));}}return cnt;}
};

DFS：

class Solution {
public:vector<vector<int> > visit;int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int r;int c;int maxAreaOfIsland(vector<vector<int> >& grid) {r = grid.size();if(r == 0)return 0;c = grid[0].size();visit = vector<vector<int> >(r, vector<int>(c, 0));int res = 0;for(int i = 0; i < r; i++){for(int j = 0; j < c; j++){if(visit[i][j] != 1 && grid[i][j] == 1){res = max(res, DFS(grid, i, j));}}}return res;}int DFS(vector<vector<int> >& grid, int x, int y){int cnt = 1;visit[x][y] = 1;for(int i = 0; i < 4; i++){int newx = x + dx[i];int newy = y + dy[i];if(newx < 0 || newx >= r || newy < 0 || newy >= c)continue;if(visit[newx][newy] == 1)continue;if(grid[newx][newy] == 0)continue;cnt += DFS(grid, newx, newy);}return cnt;}
};