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 。

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 。

输入：grid = [[0,0,0,0,0,0,0,0]]
输出：0

class Solution {public int maxAreaOfIsland(int[][] grid) {int res = 0;for (int r = 0; r < grid.length; r++) {for (int c = 0; c < grid[0].length; c++) {if (grid[r][c] == 1) {res = Math.max(res, dfs(grid, r, c));}}}return res;}private int dfs(int[][] grid, int r, int c) {if (!isGrid(grid, r, c)) {return 0;}if (grid[r][c] == 2) {return 0;}if (grid[r][c] == 0) {return 0;}grid[r][c] = 2;return 1 ++ dfs(grid, r + 1, c)+ dfs(grid, r - 1, c)+ dfs(grid, r, c + 1)+ dfs(grid, r, c - 1);}// 是否在网格区域内private boolean isGrid(int[][] grid, int r, int c) {return r >= 0 && c >= 0 && r < grid.length && c < grid[0].length;}
}

原题链接：力扣

视频讲解链接：岛屿最大面积_哔哩哔哩_bilibili