【LeetCode】695. 岛屿的最大面积

695. 岛屿的最大面积

给你一个大小为 m x n 的二进制矩阵 grid 。

岛屿 是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1 必须在 水平或者竖直的四个方向上 相邻。你可以假设 grid 的四个边缘都被 0（代表水）包围着。

岛屿的面积是岛上值为 1 的单元格的数目。

计算并返回 grid 中最大的岛屿面积。如果没有岛屿，则返回面积为 0 。

示例 1：

输入：grid = [[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]]
输出：6
解释：答案不应该是 11 ，因为岛屿只能包含水平或垂直这四个方向上的 1 。

示例 2：

输入：grid = [[0,0,0,0,0,0,0,0]]
输出：0

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• grid[i][j] 为 0 或 1

DFS

class Solution {
    
    
private:
    const int dx[4] = {
    
    0, 0, 1, -1};
    const int dy[4] = {
    
    -1, 1, 0, 0};
    int dfs(vector<vector<int>>& grid, int x, int y) {
    
            
        if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] == 0) return 0;
        grid[x][y] = 0; 
        int area = 1;
        area += dfs(grid, x, y + 1);
        area += dfs(grid, x, y - 1);
        area += dfs(grid, x - 1, y);
        area += dfs(grid, x + 1, y);
        return area;
    }

public:
    
    int maxAreaOfIsland(vector<vector<int>>& grid) {
    
    
        int maxArea = 0;
        int n = grid.size(), m = grid[0].size();
        for (int i = 0; i < n; ++i) {
    
    
            for (int j = 0; j < m; ++j) {
    
    
                if (grid[i][j] == 1) {
    
    
                    maxArea = max(maxArea, dfs(grid, i, j));
                }
            }
        }
        return maxArea;
    }
};

BFS

class Solution {
    
    
private:
    const int dx[4] = {
    
    0, 0, 1, -1};
    const int dy[4] = {
    
    -1, 1, 0, 0};
    int bfs(vector<vector<int>>& grid, int x, int y) {
    
    
        int area = 1;
        queue<pair<int, int>> pointQueue;
        pointQueue.emplace(x, y);
        grid[x][y] = 0;
        while (!pointQueue.empty()) {
    
    
            int size = pointQueue.size();
            for (int i = 0; i < size; ++i) {
    
    
                auto point = pointQueue.front();
                pointQueue.pop();
                x = point.first;
                y = point.second;
                for (int j = 0; j < 4; ++j) {
    
    
                    int nx = x + dx[j];
                    int ny = y + dy[j];
                    if (nx >= 0 && nx < grid.size() && ny >= 0 && ny < grid[0].size() && grid[nx][ny] == 1) {
    
    
                        area += bfs(grid, nx, ny);
                    } 
                }
            }
        }
        return area;
    }

public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
    
    
        int maxArea = 0;
        int n = grid.size();
        int m = grid[0].size();
        for (int i = 0; i < n; ++i) {
    
    
            for (int j = 0; j < m; ++j) {
    
    
                if (grid[i][j] == 1) {
    
    
                    maxArea = max(maxArea, bfs(grid, i, j));
                }
            }
        }
        return maxArea;
    }
};