<p align="center">
<a href="https://www.programmercarl.com/xunlian/xunlianying.html" target="_blank">
  <img src="../pics/训练营.png" width="1000"/>
</a>
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>

# 1020. 飞地的数量 

[力扣链接](https://leetcode.cn/problems/number-of-enclaves/description/)

给你一个大小为 m x n 的二进制矩阵 grid ，其中 0 表示一个海洋单元格、1 表示一个陆地单元格。

一次 移动 是指从一个陆地单元格走到另一个相邻（上、下、左、右）的陆地单元格或跨过 grid 的边界。

返回网格中 无法 在任意次数的移动中离开网格边界的陆地单元格的数量。

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20220830100710.png) 

* 输入：grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
* 输出：3
* 解释：有三个 1 被 0 包围。一个 1 没有被包围，因为它在边界上。

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20220830100742.png) 

* 输入：grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
* 输出：0
* 解释：所有 1 都在边界上或可以到达边界。 

## 思路 

本题使用dfs，bfs，并查集都是可以的。 

本题要求找到不靠边的陆地面积，那么我们只要从周边找到陆地然后 通过 dfs或者bfs 将周边靠陆地且相邻的陆地都变成海洋，然后再去重新遍历地图的时候，统计此时还剩下的陆地就可以了。

如图，在遍历地图周围四个边，靠地图四边的陆地，都为绿色，

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20220830104632.png)

在遇到地图周边陆地的时候，将1都变为0，此时地图为这样：

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20220830104651.png)

然后我们再去遍历这个地图，遇到有陆地的地方，去采用深搜或者广搜，边统计所有陆地。

如果对深搜或者广搜不够了解，建议先看这里：[深度优先搜索精讲](https://programmercarl.com/图论深搜理论基础.html)，[广度优先搜索精讲](https://programmercarl.com/图论广搜理论基础.html)。


采用深度优先搜索的代码如下： 

```CPP 
class Solution {
private:
    int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; // 保存四个方向
    int count; // 统计符合题目要求的陆地空格数量
    void dfs(vector<vector<int>>& grid, int x, int y) {
        grid[x][y] = 0;
        count++;
        for (int i = 0; i < 4; i++) { // 向四个方向遍历
            int nextx = x + dir[i][0];
            int nexty = y + dir[i][1];
            // 超过边界
            if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
            // 不符合条件，不继续遍历
            if (grid[nextx][nexty] == 0) continue;

            dfs (grid, nextx, nexty);
        }
        return;
    }

public:
    int numEnclaves(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        // 从左侧边，和右侧边 向中间遍历
        for (int i = 0; i < n; i++) {
            if (grid[i][0] == 1) dfs(grid, i, 0);
            if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
        }
        // 从上边和下边 向中间遍历
        for (int j = 0; j < m; j++) {
            if (grid[0][j] == 1) dfs(grid, 0, j);
            if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
        }
        count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) dfs(grid, i, j);
            }
        }
        return count;
    }
};
```

采用广度优先搜索的代码如下：

```CPP 
class Solution {
private:
int count = 0;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(vector<vector<int>>& grid, int x, int y) {
    queue<pair<int, int>> que;
    que.push({x, y});
    grid[x][y] = 0; // 只要加入队列，立刻标记
    count++;
    while(!que.empty()) {
        pair<int ,int> cur = que.front(); que.pop();
        int curx = cur.first;
        int cury = cur.second;
        for (int i = 0; i < 4; i++) {
            int nextx = curx + dir[i][0];
            int nexty = cury + dir[i][1];
            if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;  // 越界了，直接跳过
            if (grid[nextx][nexty] == 1) {
                que.push({nextx, nexty});
                count++;
                grid[nextx][nexty] = 0; // 只要加入队列立刻标记
            }
        }
    }

}

public:
    int numEnclaves(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        // 从左侧边，和右侧边 向中间遍历
        for (int i = 0; i < n; i++) {
            if (grid[i][0] == 1) bfs(grid, i, 0);
            if (grid[i][m - 1] == 1) bfs(grid, i, m - 1);
        }
        // 从上边和下边 向中间遍历
        for (int j = 0; j < m; j++) {
            if (grid[0][j] == 1) bfs(grid, 0, j);
            if (grid[n - 1][j] == 1) bfs(grid, n - 1, j);
        }
        count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) bfs(grid, i, j);
            }
        }
        return count;
    }
};
```
## 其他语言版本

### Java

深度优先遍历(没有终止条件 + 空間優化（淹沒島嶼，沒有使用visited數組）)
```java
//DFS
class Solution {
    int count = 0;
    int[][] dir ={
        {0, 1},
        {1, 0},
        {-1, 0},
        {0, -1}
    };
    private void dfs(int[][] grid, int x, int y){
        if(grid[x][y] == 0)
            return;
        
        grid[x][y] = 0;
        count++;

        for(int i = 0; i < 4; i++){
            int nextX = x + dir[i][0];
            int nextY = y + dir[i][1];

            if(nextX < 0 || nextY < 0 || nextX >= grid.length || nextY >= grid[0].length)
                continue;
            dfs(grid, nextX, nextY);
        }
        
    }

    public int numEnclaves(int[][] grid) {
        for(int i = 0; i < grid.length; i++){
            if(grid[i][0] == 1)
                dfs(grid, i, 0);
            if(grid[i][grid[0].length - 1] == 1)
                dfs(grid, i, grid[0].length - 1);
        }
        //初始化的時候，j 的上下限有調整過，必免重複操作。
        for(int j = 1; j < grid[0].length - 1; j++){
            if(grid[0][j] == 1)
                dfs(grid, 0, j);
            if(grid[grid.length - 1][j] == 1)
                dfs(grid, grid.length - 1, j);
        }
        count = 0;

        for(int i = 1; i < grid.length - 1; i++){
            for(int j = 1; j < grid[0].length - 1; j++){
                if(grid[i][j] == 1)
                    dfs(grid, i, j);
            }
        }
        return count;
    }
}
```

深度优先遍历(没有终止条件)

```java
class Solution {
    // 四个方向
    private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

    // 深度优先遍历，把可以通向边缘部分的 1 全部标记成 true
    public void dfs(int[][] grid, int row, int col, boolean[][] visited) {
        for (int[] current: position) {
            int newRow = row + current[0], newCol = col + current[1];
            // 下标越界直接跳过
            if (newRow < 0 || newRow >= grid.length || newCol < 0 || newCol >= grid[0].length) continue;
            // 当前位置不是 1 或者已经被访问了就直接跳过
            if (grid[newRow][newCol] != 1 || visited[newRow][newCol]) continue;
            visited[newRow][newCol] = true;
            dfs(grid, newRow, newCol, visited);
        }
    }

    public int numEnclaves(int[][] grid) {
        int rowSize = grid.length, colSize = grid[0].length, ans = 0;	// ans 记录答案
        // 标记数组记录每个值为 1 的位置是否可以到达边界，可以为 true，反之为 false
        boolean[][] visited = new boolean[rowSize][colSize];
        // 左侧边界和右侧边界查找 1 进行标记并进行深度优先遍历
        for (int row = 0; row < rowSize; row++) {
            if (grid[row][0] == 1 && !visited[row][0]) {
                visited[row][0] = true;
                dfs(grid, row, 0, visited);
            }
            if (grid[row][colSize - 1] == 1 && !visited[row][colSize - 1]) {
                visited[row][colSize - 1] = true;
                dfs(grid, row, colSize - 1, visited);
            }
        }
        // 上边界和下边界遍历，但是四个角不用遍历，因为上面已经遍历到了
        for (int col = 1; col < colSize - 1; col++) {
            if (grid[0][col] == 1 && !visited[0][col]) {
                visited[0][col] = true;
                dfs(grid, 0, col, visited);
            }
            if (grid[rowSize - 1][col] == 1 && !visited[rowSize - 1][col]) {
                visited[rowSize - 1][col] = true;
                dfs(grid, rowSize - 1, col, visited);
            }
        }
        // 查找没有标记过的 1，记录到 ans 中
        for (int row = 0; row < rowSize; row++) {
            for (int col = 0; col < colSize; col++) {
                if (grid[row][col] == 1 && !visited[row][col]) ++ans;
            }
        }
        return ans;
    }
}
```

广度优先遍历（使用visited數組）

```java
class Solution {
    // 四个方向
    private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

    // 广度优先遍历，把可以通向边缘部分的 1 全部标记成 true
    public void bfs(int[][] grid, Queue<int[]> queue, boolean[][] visited) {
        while (!queue.isEmpty()) {
            int[] curPos = queue.poll();
            for (int[] current: position) {
                int row = curPos[0] + current[0], col = curPos[1] + current[1];
                // 下标越界直接跳过
                if (row < 0 || row >= grid.length || col < 0 || col >= grid[0].length) 
                    continue;
                // 当前位置不是 1 或者已经被访问了就直接跳过
                if (visited[row][col] || grid[row][col] == 0) continue;
                visited[row][col] = true;
                queue.add(new int[]{row, col});
            }
        }
    }

    public int numEnclaves(int[][] grid) {
        int rowSize = grid.length, colSize = grid[0].length, ans = 0;	// ans 记录答案
        // 标记数组记录每个值为 1 的位置是否可以到达边界，可以为 true，反之为 false
        boolean[][] visited = new boolean[rowSize][colSize];
        Queue<int[]> queue = new ArrayDeque<>();
        // 搜索左侧边界和右侧边界查找 1 存入队列
        for (int row = 0; row < rowSize; row++) {
            if (grid[row][0] == 1) {
                visited[row][0] = true;
                queue.add(new int[]{row, 0});
            }
            if (grid[row][colSize - 1] == 1) {
                visited[row][colSize - 1] = true;
                queue.add(new int[]{row, colSize - 1});
            }
        }
        // 搜索上边界和下边界遍历，但是四个角不用遍历，因为上面已经遍历到了
        for (int col = 1; col < colSize - 1; col++) {
            if (grid[0][col] == 1) {
                visited[0][col] = true;
                queue.add(new int[]{0, col});
            }
            if (grid[rowSize - 1][col] == 1 && !visited[rowSize - 1][col]) {
                visited[rowSize - 1][col] = true;
                queue.add(new int[]{rowSize - 1, col});
            }
        }
        bfs(grid, queue, visited);		// 广度优先遍历
        // 查找没有标记过的 1，记录到 ans 中
        for (int row = 0; row < rowSize; row++) {
            for (int col = 0; col < colSize; col++) {
                if (grid[row][col] == 1 && !visited[row][col]) ++ans;
            }
        }
        return ans;
    }
}
```

廣度优先遍历(空間優化（淹沒島嶼，沒有使用visited數組）)
```java
//BFS
class Solution {
    int count = 0;
    int[][] dir ={
        {0, 1},
        {1, 0},
        {-1, 0},
        {0, -1}
    };
    private void bfs(int[][] grid, int x, int y){
        Queue<Integer> que = new LinkedList<>();
        que.offer(x);
        que.offer(y);
        count++;
        grid[x][y] = 0;
        
        while(!que.isEmpty()){
            int currX = que.poll();
            int currY = que.poll();

            for(int i = 0; i < 4; i++){
                int nextX = currX + dir[i][0];
                int nextY = currY + dir[i][1];

                if(nextX < 0 || nextY < 0 || nextX >= grid.length || nextY >= grid[0].length)
                    continue;

                if(grid[nextX][nextY] == 1){
                    que.offer(nextX);
                    que.offer(nextY);
                    count++;
                    grid[nextX][nextY] = 0;
                }
            }
        }
    }

    public int numEnclaves(int[][] grid) {
        for(int i = 0; i < grid.length; i++){
            if(grid[i][0] == 1)
                bfs(grid, i, 0);
            if(grid[i][grid[0].length - 1] == 1)
                bfs(grid, i, grid[0].length - 1);
        }
        for(int j = 1; j < grid[0].length; j++){
            if(grid[0][j] == 1)
                bfs(grid, 0 , j);
            if(grid[grid.length - 1][j] == 1)
                bfs(grid, grid.length - 1, j);
        }
        count = 0;
        for(int i = 1; i < grid.length - 1; i++){
            for(int j = 1; j < grid[0].length - 1; j++){
                if(grid[i][j] == 1)
                    bfs(grid,i ,j);
            }
        }
        return count;
    }
}


```

### Python

深度优先遍历

```Python
class Solution:
    def __init__(self):
        self.position = [[-1, 0], [0, 1], [1, 0], [0, -1]]	# 四个方向

    # 深度优先遍历，把可以通向边缘部分的 1 全部标记成 true
    def dfs(self, grid: List[List[int]], row: int, col: int, visited: List[List[bool]]) -> None:
        for current in self.position:
            newRow, newCol = row + current[0], col + current[1]
            # 索引下标越界
            if newRow < 0 or newRow >= len(grid) or newCol < 0 or newCol >= len(grid[0]): 
                continue
            # 当前位置值不是 1 或者已经被访问过了
            if grid[newRow][newCol] == 0 or visited[newRow][newCol]: continue
            visited[newRow][newCol] = True
            self.dfs(grid, newRow, newCol, visited)

    def numEnclaves(self, grid: List[List[int]]) -> int:
        rowSize, colSize, ans = len(grid), len(grid[0]), 0
        # 标记数组记录每个值为 1 的位置是否可以到达边界，可以为 True，反之为 False
        visited = [[False for _ in range(colSize)] for _ in range(rowSize)]
        # 搜索左边界和右边界，对值为 1 的位置进行深度优先遍历
        for row in range(rowSize):
            if grid[row][0] == 1:
                visited[row][0] = True
                self.dfs(grid, row, 0, visited)
            if grid[row][colSize - 1] == 1:
                visited[row][colSize - 1] = True
                self.dfs(grid, row, colSize - 1, visited)
        # 搜索上边界和下边界，对值为 1 的位置进行深度优先遍历，但是四个角不需要，因为上面遍历过了
        for col in range(1, colSize - 1):
            if grid[0][col] == 1:
                visited[0][col] = True
                self.dfs(grid, 0, col, visited)
            if grid[rowSize - 1][col] == 1:
                visited[rowSize - 1][col] = True
                self.dfs(grid, rowSize - 1, col, visited)
        # 找出矩阵中值为 1 但是没有被标记过的位置，记录答案
        for row in range(rowSize):
            for col in range(colSize):
                if grid[row][col] == 1 and not visited[row][col]:
                    ans += 1
        return ans
```

广度优先遍历

```Python
class Solution:
    def __init__(self):
        self.position = [[-1, 0], [0, 1], [1, 0], [0, -1]]	# 四个方向

    # 广度优先遍历，把可以通向边缘部分的 1 全部标记成 true
    def bfs(self, grid: List[List[int]], queue: deque, visited: List[List[bool]]) -> None:
        while queue:
            curPos = queue.popleft()
            for current in self.position:
                row, col = curPos[0] + current[0], curPos[1] + current[1]
                # 索引下标越界
                if row < 0 or row >= len(grid) or col < 0 or col >= len(grid[0]): continue
                # 当前位置值不是 1 或者已经被访问过了
                if grid[row][col] == 0 or visited[row][col]: continue
                visited[row][col] = True
                queue.append([row, col])


    def numEnclaves(self, grid: List[List[int]]) -> int:
        rowSize, colSize, ans = len(grid), len(grid[0]), 0
        # 标记数组记录每个值为 1 的位置是否可以到达边界，可以为 True，反之为 False
        visited = [[False for _ in range(colSize)] for _ in range(rowSize)]
        queue = deque()		# 队列
        # 搜索左侧边界和右侧边界查找 1 存入队列
        for row in range(rowSize):
            if grid[row][0] == 1:
                visited[row][0] = True
                queue.append([row, 0])
            if grid[row][colSize - 1] == 1:
                visited[row][colSize - 1] = True
                queue.append([row, colSize - 1])
        # 搜索上边界和下边界查找 1 存入队列，但是四个角不用遍历，因为上面已经遍历到了
        for col in range(1, colSize - 1):
            if grid[0][col] == 1:
                visited[0][col] = True
                queue.append([0, col])
            if grid[rowSize - 1][col] == 1:
                visited[rowSize - 1][col] = True
                queue.append([rowSize - 1, col])
        self.bfs(grid, queue, visited)	# 广度优先遍历
        # 找出矩阵中值为 1 但是没有被标记过的位置，记录答案
        for row in range(rowSize):
            for col in range(colSize):
                if grid[row][col] == 1 and not visited[row][col]:
                    ans += 1
        return ans
```

### Go

dfs:

```go
var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
var count int = 0

func numEnclaves(grid [][]int) int {
	rows, cols := len(grid), len(grid[0])
	// 行
	for i := range grid[0] {
		if grid[0][i] == 1 {
			dfs(grid, 0, i)
		}
		if grid[rows-1][i] == 1 {
			dfs(grid, rows-1, i)
		}
	}
	// 列
	for j := range grid {
		if grid[j][0] == 1 {
			dfs(grid, j, 0)
		}
		if grid[j][cols-1] == 1 {
			dfs(grid, j, cols-1)
		}
	}
	count = 0
	for i := range grid {
		for j := range grid[0] {
			if grid[i][j] == 1 {
				dfs(grid, i, j)
			}
		}
	}
	return count
}

func dfs(grid [][]int, i, j int) {
	grid[i][j] = 0
	count++
	for _, d := range DIRECTIONS {
		x, y := i+d[0], j+d[1]
		if x < 0 || x >= len(grid) || y < 0 || y >= len(grid[0]) {
			continue
		}
		if grid[x][y] == 1 {
			dfs(grid, x, y)
		}
	}
}
```

bfs:

```go
var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
var count int = 0

func numEnclaves(grid [][]int) int {
	rows, cols := len(grid), len(grid[0])
	// 行
	for i := range grid[0] {
		if grid[0][i] == 1 {
			bfs(grid, 0, i)
		}
		if grid[rows-1][i] == 1 {
			bfs(grid, rows-1, i)
		}
	}
	// 列
	for j := range grid {
		if grid[j][0] == 1 {
			bfs(grid, j, 0)
		}
		if grid[j][cols-1] == 1 {
			bfs(grid, j, cols-1)
		}
	}
	count = 0
	for i := range grid {
		for j := range grid[0] {
			if grid[i][j] == 1 {
				bfs(grid, i, j)
			}
		}
	}
	return count
}

func bfs(grid [][]int, i, j int) {
	queue := [][]int{}
	queue = append(queue, []int{i, j})
	grid[i][j] = 0
	count++
	for len(queue) > 0 {
		cur := queue[0]
		queue = queue[1:]
		for _, d := range DIRECTIONS {
			x, y := cur[0]+d[0], cur[1]+d[1]
			if x < 0 || x >= len(grid) || y < 0 || y >= len(grid[0]) {
				continue
			}
			if grid[x][y] == 1 {
				count++
				queue = append(queue, []int{x, y})
				grid[x][y] = 0
			}
		}
	}
}
```

### JavaScript

```js
/**
 * @param {number[][]} grid
 * @return {number}
 */
var numEnclaves = function (grid) {
  let row = grid.length;
  let col = grid[0].length;
  let count = 0;

  // Check the first and last row, if there is a 1, then change all the connected 1s to 0 and don't count them.
  for (let j = 0; j < col; j++) {
    if (grid[0][j] === 1) {
      dfs(0, j, false);
    }
    if (grid[row - 1][j] === 1) {
      dfs(row - 1, j, false);
    }
  }

  // Check the first and last column, if there is a 1, then change all the connected 1s to 0 and don't count them.
  for (let i = 0; i < row; i++) {
    if (grid[i][0] === 1) {
      dfs(i, 0, false);
    }
    if (grid[i][col - 1] === 1) {
      dfs(i, col - 1, false);
    }
  }

  // Check the rest of the grid, if there is a 1, then change all the connected 1s to 0 and count them.
  for (let i = 1; i < row - 1; i++) {
    for (let j = 1; j < col - 1; j++) {
      dfs(i, j, true);
    }
  }

  function dfs(i, j, isCounting) {
    let condition = i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] === 0;

    if (condition) return;
    if (isCounting) count++;

    grid[i][j] = 0;

    dfs(i - 1, j, isCounting);
    dfs(i + 1, j, isCounting);
    dfs(i, j - 1, isCounting);
    dfs(i, j + 1, isCounting);
  }

  return count;
};
```

### Rust

dfs:

```rust
impl Solution {
    const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)];

    pub fn num_enclaves(mut grid: Vec<Vec<i32>>) -> i32 {
        for i in 0..grid.len() {
            for j in 0..grid[0].len() {
                if (i == 0 || i == grid.len() - 1 || j == 0 || j == grid[0].len() - 1)
                    && grid[i][j] == 1
                {
                    Self::dfs(&mut grid, (i as i32, j as i32));
                }
            }
        }
        grid.iter()
            .map(|nums| nums.iter().filter(|&&num| num == 1).count() as i32)
            .sum()
    }

    pub fn dfs(grid: &mut [Vec<i32>], (x, y): (i32, i32)) {
        grid[x as usize][y as usize] = 0;
        for (dx, dy) in Self::DIRECTIONS {
            let (nx, ny) = (x + dx, y + dy);
            if nx < 0 || nx >= grid.len() as i32 || ny < 0 || ny >= grid[0].len() as i32 {
                continue;
            }
            if grid[nx as usize][ny as usize] == 0 {
                continue;
            }
            Self::dfs(grid, (nx, ny));
        }
    }
}
```

bfs:

```rust
use std::collections::VecDeque;
impl Solution {
    const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)];

    pub fn num_enclaves(mut grid: Vec<Vec<i32>>) -> i32 {
        for i in 0..grid.len() {
            for j in 0..grid[0].len() {
                if (i == 0 || i == grid.len() - 1 || j == 0 || j == grid[0].len() - 1)
                    && grid[i][j] == 1
                {
                    // Self::dfs(&mut grid, (i as i32, j as i32));
                    Self::bfs(&mut grid, (i as i32, j as i32));
                }
            }
        }
        grid.iter()
            .map(|nums| nums.iter().filter(|&&num| num == 1).count() as i32)
            .sum()
    }

    pub fn bfs(grid: &mut [Vec<i32>], (x, y): (i32, i32)) {
        let mut queue = VecDeque::new();
        queue.push_back((x, y));
        grid[x as usize][y as usize] = 0;
        while let Some((cur_x, cur_y)) = queue.pop_front() {
            for (dx, dy) in Self::DIRECTIONS {
                let (nx, ny) = (cur_x + dx, cur_y + dy);
                if nx < 0 || nx >= grid.len() as i32 || ny < 0 || ny >= grid[0].len() as i32 {
                    continue;
                }

                if grid[nx as usize][ny as usize] == 0 {
                    continue;
                }
                queue.push_back((nx, ny));
                grid[nx as usize][ny as usize] = 0;
            }
        }
    }
}
```


## 类似题目 

* 1254. 统计封闭岛屿的数目  



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