跳转至

1020. 飞地的数量

题目描述

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。

 

示例 1：

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。

示例 2：

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。

 

提示：

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 500
  • grid[i][j] 的值为 0 或 1

解法

方法一：DFS

我们可以从边界上的陆地开始进行深度优先搜索，将所有与边界相连的陆地都标记为 $0$。最后，统计剩余的 $1$ 的个数，即为答案。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        def dfs(i: int, j: int):
            grid[i][j] = 0
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[x][y]:
                    dfs(x, y)

        m, n = len(grid), len(grid[0])
        dirs = (-1, 0, 1, 0, -1)
        for j in range(n):
            for i in (0, m - 1):
                if grid[i][j]:
                    dfs(i, j)
        for i in range(m):
            for j in (0, n - 1):
                if grid[i][j]:
                    dfs(i, j)
        return sum(sum(row) for row in grid)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
class Solution {
    private int[][] grid;

    public int numEnclaves(int[][] grid) {
        this.grid = grid;
        int m = grid.length, n = grid[0].length;
        for (int j = 0; j < n; j++) {
            for (int i : List.of(0, m - 1)) {
                if (grid[i][j] == 1) {
                    dfs(i, j);
                }
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j : List.of(0, n - 1)) {
                if (grid[i][j] == 1) {
                    dfs(i, j);
                }
            }
        }
        int ans = 0;
        for (var row : grid) {
            for (int x : row) {
                ans += x;
            }
        }
        return ans;
    }

    private void dfs(int i, int j) {
        grid[i][j] = 0;
        final int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; k++) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < grid.length && y >= 0 && y < grid[0].length && grid[x][y] == 1) {
                dfs(x, y);
            }
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        const int dirs[5] = {-1, 0, 1, 0, -1};
        function<void(int, int)> dfs = [&](int i, int j) {
            grid[i][j] = 0;
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                    dfs(x, y);
                }
            }
        };
        for (int j = 0; j < n; ++j) {
            for (int i : {0, m - 1}) {
                if (grid[i][j] == 1) {
                    dfs(i, j);
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j : {0, n - 1}) {
                if (grid[i][j] == 1) {
                    dfs(i, j);
                }
            }
        }
        int ans = 0;
        for (const auto& row : grid) {
            ans += accumulate(row.begin(), row.end(), 0);
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
func numEnclaves(grid [][]int) (ans int) {
    m, n := len(grid), len(grid[0])
    dirs := [5]int{-1, 0, 1, 0, -1}
    var dfs func(i, j int)
    dfs = func(i, j int) {
        grid[i][j] = 0
        for k := 0; k < 4; k++ {
            x, y := i+dirs[k], j+dirs[k+1]
            if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 {
                dfs(x, y)
            }
        }
    }
    for j := 0; j < n; j++ {
        for _, i := range [2]int{0, m - 1} {
            if grid[i][j] == 1 {
                dfs(i, j)
            }
        }
    }
    for i := 0; i < m; i++ {
        for _, j := range [2]int{0, n - 1} {
            if grid[i][j] == 1 {
                dfs(i, j)
            }
        }
    }
    for _, row := range grid {
        for _, x := range row {
            ans += x
        }
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
function numEnclaves(grid: number[][]): number {
    const [m, n] = [grid.length, grid[0].length];
    const dirs: number[] = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number) => {
        grid[i][j] = 0;
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y <= n && grid[x][y] === 1) {
                dfs(x, y);
            }
        }
    };
    for (let j = 0; j < n; ++j) {
        for (let i of [0, m - 1]) {
            if (grid[i][j] === 1) {
                dfs(i, j);
            }
        }
    }
    for (let i = 0; i < m; ++i) {
        for (let j of [0, n - 1]) {
            if (grid[i][j] === 1) {
                dfs(i, j);
            }
        }
    }
    return grid.flat().reduce((acc, cur) => acc + cur, 0);
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
impl Solution {
    pub fn num_enclaves(mut grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let dirs = [-1, 0, 1, 0, -1];

        fn dfs(grid: &mut Vec<Vec<i32>>, i: usize, j: usize, dirs: &[i32; 5]) {
            grid[i][j] = 0;
            for k in 0..4 {
                let (x, y) = (i as i32 + dirs[k], j as i32 + dirs[k + 1]);
                if let Some(row) = grid.get_mut(x as usize) {
                    if let Some(&mut 1) = row.get_mut(y as usize) {
                        dfs(grid, x as usize, y as usize, dirs);
                    }
                }
            }
        }

        for j in 0..n {
            if grid[0][j] == 1 {
                dfs(&mut grid, 0, j, &dirs);
            }
            if grid[m - 1][j] == 1 {
                dfs(&mut grid, m - 1, j, &dirs);
            }
        }

        for i in 0..m {
            if grid[i][0] == 1 {
                dfs(&mut grid, i, 0, &dirs);
            }
            if grid[i][n - 1] == 1 {
                dfs(&mut grid, i, n - 1, &dirs);
            }
        }

        grid.into_iter().flatten().filter(|&x| x == 1).count() as i32
    }
}

方法二：BFS

我们也可以使用广度优先搜索的方法，将边界上的陆地入队，然后进行广度优先搜索，将所有与边界相连的陆地都标记为 $0$。最后，统计剩余的 $1$ 的个数，即为答案。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        q = deque()
        for j in range(n):
            for i in (0, m - 1):
                if grid[i][j]:
                    q.append((i, j))
                    grid[i][j] = 0
        for i in range(m):
            for j in (0, n - 1):
                if grid[i][j]:
                    q.append((i, j))
                    grid[i][j] = 0
        dirs = (-1, 0, 1, 0, -1)
        while q:
            i, j = q.popleft()
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[x][y]:
                    q.append((x, y))
                    grid[x][y] = 0
        return sum(sum(row) for row in grid)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
class Solution {
    public int numEnclaves(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        Deque<int[]> q = new ArrayDeque<>();
        for (int j = 0; j < n; ++j) {
            for (int i : List.of(0, m - 1)) {
                if (grid[i][j] == 1) {
                    q.offer(new int[] {i, j});
                    grid[i][j] = 0;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j : List.of(0, n - 1)) {
                if (grid[i][j] == 1) {
                    q.offer(new int[] {i, j});
                    grid[i][j] = 0;
                }
            }
        }
        final int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            int[] p = q.poll();
            int i = p[0], j = p[1];
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                    q.offer(new int[] {x, y});
                    grid[x][y] = 0;
                }
            }
        }
        int ans = 0;
        for (var row : grid) {
            for (int x : row) {
                ans += x;
            }
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int dirs[5] = {-1, 0, 1, 0, -1};
        queue<pair<int, int>> q;
        for (int j = 0; j < n; ++j) {
            for (int i : {0, m - 1}) {
                if (grid[i][j] == 1) {
                    q.emplace(i, j);
                    grid[i][j] = 0;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j : {0, n - 1}) {
                if (grid[i][j] == 1) {
                    q.emplace(i, j);
                    grid[i][j] = 0;
                }
            }
        }
        while (!q.empty()) {
            auto [i, j] = q.front();
            q.pop();
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                    q.emplace(x, y);
                    grid[x][y] = 0;
                }
            }
        }
        int ans = 0;
        for (const auto& row : grid) {
            ans += accumulate(row.begin(), row.end(), 0);
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
func numEnclaves(grid [][]int) (ans int) {
    m, n := len(grid), len(grid[0])
    dirs := [5]int{-1, 0, 1, 0, -1}
    q := [][2]int{}
    for j := 0; j < n; j++ {
        for _, i := range []int{0, m - 1} {
            if grid[i][j] == 1 {
                q = append(q, [2]int{i, j})
                grid[i][j] = 0
            }
        }
    }
    for i := 0; i < m; i++ {
        for _, j := range []int{0, n - 1} {
            if grid[i][j] == 1 {
                q = append(q, [2]int{i, j})
                grid[i][j] = 0
            }
        }
    }
    for len(q) > 0 {
        i, j := q[0][0], q[0][1]
        q = q[1:]
        for k := 0; k < 4; k++ {
            x, y := i+dirs[k], j+dirs[k+1]
            if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 {
                q = append(q, [2]int{x, y})
                grid[x][y] = 0
            }
        }
    }
    for _, row := range grid {
        for _, x := range row {
            ans += x
        }
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
function numEnclaves(grid: number[][]): number {
    const [m, n] = [grid.length, grid[0].length];
    const dirs = [-1, 0, 1, 0, -1];
    const q: number[][] = [];
    for (let j = 0; j < n; ++j) {
        for (let i of [0, m - 1]) {
            if (grid[i][j] === 1) {
                q.push([i, j]);
                grid[i][j] = 0;
            }
        }
    }
    for (let i = 0; i < m; ++i) {
        for (let j of [0, n - 1]) {
            if (grid[i][j] === 1) {
                q.push([i, j]);
                grid[i][j] = 0;
            }
        }
    }
    while (q.length) {
        const [i, j] = q.pop()!;
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] === 1) {
                q.push([x, y]);
                grid[x][y] = 0;
            }
        }
    }
    return grid.flat().reduce((acc, cur) => acc + cur, 0);
}

评论