LWC 53：695. Max Area of Island

LWC 53：695. Max Area of Island

传送门：695. Max Area of Island

Problem:

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.

思路：
DFS求出邻接的面积，更新每个区域的最大值即可。

代码如下：

    int[][] dir = {{1, 0},{-1, 0},{0, 1},{0, -1}};    public int maxAreaOfIsland(int[][] grid) {        int n = grid.length;        if (n == 0) return 0;        int m = grid[0].length;        if (m == 0) return 0;        int max = 0;        for (int i = 0; i < n; ++i) {            for (int j = 0; j < m; ++j) {                if (grid[i][j] == 1) {                    max = Math.max(max, dfs(grid, i, j, n, m, new boolean[n][m]));                }            }        }        return max;    }    public int dfs(int[][] grid, int i, int j, int n, int m, boolean[][] vis) {        int res = 1;        vis[i][j] = true;        for (int[] d : dir) {            int nx = i + d[0];            int ny = j + d[1];            if (nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny] && grid[nx][ny] == 1) {                res += dfs(grid, nx, ny, n, m, vis);            }        }        return res;    }