# 1020. Number of Enclaves (M)

You are given an `m x n` binary matrix `grid`, where `0` represents a sea cell and `1` represents a land cell.

A **move** consists of walking from one land cell to another adjacent (**4-directionally**) land cell or walking off the boundary of the `grid`.

Return *the number of land cells in* `grid` *for which we cannot walk off the boundary of the grid in any number of **moves***.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/02/18/enclaves1.jpg)

```
Input: grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 3
Explanation: There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/02/18/enclaves2.jpg)

```
Input: grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 0
Explanation: All 1s are either on the boundary or can reach the boundary.
```

 

**Constraints:**

* `m == grid.length`
* `n == grid[i].length`
* `1 <= m, n <= 500`
* `grid[i][j]` is either `0` or `1`.

### **Solution:**

Similar as LeetCode 1254

这题不让你求封闭岛屿的数量，而是求封闭岛屿的面积总和。

其实思路都是一样的，先把靠边的陆地淹掉，然后去数剩下的陆地数量就行了.

```
class Solution {
    public int numEnclaves(int[][] grid) {
        
        int m = grid.length;
        int n = grid[0].length;
        for(int i = 0; i< n; i++)
        {
            dfs(grid, 0, i);
            dfs(grid, m-1, i);
        }
        for(int i = 0; i< m; i++)
        {
            dfs(grid, i, 0);
            dfs(grid,i, n-1);
        }
        int result = 0;
        for(int i = 0; i< m ;i++)
        {
            for(int j = 0; j< n; j++)
            {
                if(grid[i][j] == 1)
                {
                    result++;
                }
            }
        }
        return result;
        
    }
    
    public void dfs(int[][] grid, int x, int y)
    {
        int m = grid.length;
        int n = grid[0].length;
        if(x<0 || y<0 || x>=m || y>=n) return;
        if(grid[x][y] == 0) return;
        grid[x][y] = 0;
        dfs(grid, x-1, y);
        dfs(grid, x+1, y);
        dfs(grid, x, y-1);
        dfs(grid, x, y+1);
    }
}
```


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