> For the complete documentation index, see [llms.txt](https://junnie.gitbook.io/nine-chapter/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://junnie.gitbook.io/nine-chapter/4.depth-first-search/dao-yu-wen-ti/695.-max-area-of-island-m.md).

# 695. Max Area of Island (M)

You are given an `m x n` binary matrix `grid`. 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.

The **area** of an island is the number of cells with a value `1` in the island.

Return *the maximum **area** of an island in* `grid`. If there is no island, return `0`.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/05/01/maxarea1-grid.jpg)

```
Input: 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]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.
```

**Example 2:**

```
Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0
```

 

**Constraints:**

* `m == grid.length`
* `n == grid[i].length`
* `1 <= m, n <= 50`
* `grid[i][j]` is either `0` or `1`.

### **Solution:**

**这题的大体思路和之前完全一样，只不过 `dfs` 函数淹没岛屿的同时，还应该想办法记录这个岛屿的面积**。

我们可以给 `dfs` 函数设置返回值，记录每次淹没的陆地的个数，直接看解法吧：

```java
int maxAreaOfIsland(int[][] grid) {
    // 记录岛屿的最大面积
    int res = 0;
    int m = grid.length, n = grid[0].length;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid[i][j] == 1) {
                // 淹没岛屿，并更新最大岛屿面积
                res = Math.max(res, dfs(grid, i, j));
            }
        }
    }
    return res;
}

// 淹没与 (i, j) 相邻的陆地，并返回淹没的陆地面积
int dfs(int[][] grid, int i, int j) {
    int m = grid.length, n = grid[0].length;
    if (i < 0 || j < 0 || i >= m || j >= n) {
        // 超出索引边界
        return 0;
    }
    if (grid[i][j] == 0) {
        // 已经是海水了
        return 0;
    }
    // 将 (i, j) 变成海水
    grid[i][j] = 0;

    return dfs(grid, i + 1, j)
         + dfs(grid, i, j + 1)
         + dfs(grid, i - 1, j)
         + dfs(grid, i, j - 1) + 1;
}
```

解法和之前相比差不多，