Leetcode695.Max Area of Island岛屿的最大面积

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;
    }
};