Question

1020. 飞地的数量

English Version

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。

class Solution:
  def numEnclaves(self, grid: List[List[int]]) -> int:
    def dfs(i, j):
      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 i in range(m):
      for j in range(n):
        if grid[i][j] and (i == 0 or i == m - 1 or j == 0 or j == n - 1):
          dfs(i, j)
    return sum(v for row in grid for v in row)

class Solution {
  private int m;
  private int n;
  private int[][] grid;

  public int numEnclaves(int[][] grid) {
    this.grid = grid;
    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 && (i == 0 || i == m - 1 || j == 0 || j == n - 1)) {
          dfs(i, j);
        }
      }
    }
    int ans = 0;
    for (var row : grid) {
      for (var v : row) {
        ans += v;
      }
    }
    return ans;
  }

  private void dfs(int i, int j) {
    grid[i][j] = 0;
    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 < m && y >= 0 && y < n && grid[x][y] == 1) {
        dfs(x, y);
      }
    }
  }
}

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};
    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]) {
          dfs(x, y);
        }
      }
    };
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] && (i == 0 || i == m - 1 || j == 0 || j == n - 1)) {
          dfs(i, j);
        }
      }
    }
    int ans = 0;
    for (auto& row : grid) {
      for (auto& v : row) {
        ans += v;
      }
    }
    return ans;
  }
};

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 i, row := range grid {
    for j, v := range row {
      if v == 1 && (i == 0 || i == m-1 || j == 0 || j == n-1) {
        dfs(i, j)
      }
    }
  }
  for _, row := range grid {
    for _, v := range row {
      ans += v
    }
  }
  return
}

function numEnclaves(grid: number[][]): number {
  const m = grid.length;
  const n = grid[0].length;
  const dirs = [-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 i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      if (grid[i][j] === 1 && (i === 0 || i === m - 1 || j === 0 || j === n - 1)) {
        dfs(i, j);
      }
    }
  }
  let ans = 0;
  for (const row of grid) {
    for (const v of row) {
      ans += v;
    }
  }
  return ans;
}

impl Solution {
  fn dfs(grid: &mut Vec<Vec<i32>>, y: usize, x: usize) {
    if y >= grid.len() || x >= grid[0].len() || grid[y][x] == 0 {
      return;
    }
    grid[y][x] = 0;
    Solution::dfs(grid, y + 1, x);
    Solution::dfs(grid, y, x + 1);
    if y != 0 {
      Solution::dfs(grid, y - 1, x);
    }
    if x != 0 {
      Solution::dfs(grid, y, x - 1);
    }
  }
  pub fn num_enclaves(mut grid: Vec<Vec<i32>>) -> i32 {
    let mut res = 0;
    let m = grid.len();
    let n = grid[0].len();
    for i in 0..m {
      Solution::dfs(&mut grid, i, 0);
      Solution::dfs(&mut grid, i, n - 1);
    }
    for i in 0..n {
      Solution::dfs(&mut grid, 0, i);
      Solution::dfs(&mut grid, m - 1, i);
    }
    for i in 1..m - 1 {
      for j in 1..n - 1 {
        if grid[i][j] == 1 {
          res += 1;
        }
      }
    }
    res
  }
}

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。

class Solution:
  def numEnclaves(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    q = deque()
    for i in range(m):
      for j in range(n):
        if grid[i][j] and (i == 0 or i == m - 1 or j == 0 or j == n - 1):
          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 x >= 0 and x < m and y >= 0 and y < n and grid[x][y]:
          q.append((x, y))
          grid[x][y] = 0
    return sum(v for row in grid for v in row)

class Solution {
  public int numEnclaves(int[][] grid) {
    int m = grid.length;
    int n = grid[0].length;
    Deque<int[]> q = new ArrayDeque<>();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 1 && (i == 0 || i == m - 1 || j == 0 || j == n - 1)) {
          q.offer(new int[] {i, j});
          grid[i][j] = 0;
        }
      }
    }
    int[] dirs = {-1, 0, 1, 0, -1};
    while (!q.isEmpty()) {
      var p = q.poll();
      for (int k = 0; k < 4; ++k) {
        int x = p[0] + dirs[k], y = p[1] + 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 (var v : row) {
        ans += v;
      }
    }
    return ans;
  }
}

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 i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] && (i == 0 || i == m - 1 || j == 0 || j == n - 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]) {
          q.emplace(x, y);
          grid[x][y] = 0;
        }
      }
    }
    int ans = 0;
    for (auto& row : grid) {
      for (auto& v : row) {
        ans += v;
      }
    }
    return ans;
  }
};

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 i, row := range grid {
    for j, v := range row {
      if v == 1 && (i == 0 || i == m-1 || j == 0 || j == n-1) {
        q = append(q, [2]int{i, j})
        grid[i][j] = 0
      }
    }
  }
  for len(q) > 0 {
    p := q[0]
    q = q[1:]
    for k := 0; k < 4; k++ {
      x, y := p[0]+dirs[k], p[1]+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 _, v := range row {
      ans += v
    }
  }
  return
}

function numEnclaves(grid: number[][]): number {
  const m = grid.length;
  const n = grid[0].length;
  const dirs = [-1, 0, 1, 0, -1];
  const q: number[][] = [];
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      if (grid[i][j] === 1 && (i === 0 || i === m - 1 || j === 0 || j === n - 1)) {
        q.push([i, j]);
        grid[i][j] = 0;
      }
    }
  }
  while (q.length) {
    const [i, j] = q.shift()!;
    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;
      }
    }
  }
  let ans = 0;
  for (const row of grid) {
    for (const v of row) {
      ans += v;
    }
  }
  return ans;
}

方法三：并查集

我们还可以利用并查集的方法，将边界上的陆地与一个虚拟的节点 $(m, n)$ 进行合并，然后遍历矩阵中的所有陆地，将其与上下左右的陆地进行合并。最后，统计所有与虚拟节点 $(m, n)$ 不连通的陆地的个数，即为答案。

时间复杂度 $O(m \times n \alpha(m \times n))$，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，而 $\alpha$ 为 阿克曼函数 的反函数。

class UnionFind:
  def __init__(self, n):
    self.p = list(range(n))
    self.size = [1] * n

  def find(self, x):
    if self.p[x] != x:
      self.p[x] = self.find(self.p[x])
    return self.p[x]

  def union(self, a, b):
    pa, pb = self.find(a), self.find(b)
    if pa != pb:
      if self.size[pa] > self.size[pb]:
        self.p[pb] = pa
        self.size[pa] += self.size[pb]
      else:
        self.p[pa] = pb
        self.size[pb] += self.size[pa]


class Solution:
  def numEnclaves(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    uf = UnionFind(m * n + 1)
    dirs = (-1, 0, 1, 0, -1)
    for i, row in enumerate(grid):
      for j, v in enumerate(row):
        if v:
          if i == 0 or i == m - 1 or j == 0 or j == n - 1:
            uf.union(i * n + j, m * n)
          else:
            for a, b in pairwise(dirs):
              x, y = i + a, j + b
              if x >= 0 and x < m and y >= 0 and y < n and grid[x][y]:
                uf.union(i * n + j, x * n + y)
    return sum(
      grid[i][j] == 1 and uf.find(i * n + j) != uf.find(m * n)
      for i in range(m)
      for j in range(n)
    )

class UnionFind {
  private int[] p;
  private int[] size;

  public UnionFind(int n) {
    p = new int[n];
    size = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
      size[i] = 1;
    }
  }

  public int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }

  public void union(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa != pb) {
      if (size[pa] > size[pb]) {
        p[pb] = pa;
        size[pa] += size[pb];
      } else {
        p[pa] = pb;
        size[pb] += size[pa];
      }
    }
  }
}

class Solution {
  public int numEnclaves(int[][] grid) {
    int m = grid.length;
    int n = grid[0].length;
    UnionFind uf = new UnionFind(m * n + 1);
    int[] dirs = {-1, 0, 1, 0, -1};
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 1) {
          if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
            uf.union(i * n + j, m * n);
          } else {
            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) {
                uf.union(i * n + j, x * n + y);
              }
            }
          }
        }
      }
    }
    int ans = 0;
    for (int i = 1; i < m - 1; ++i) {
      for (int j = 1; j < n - 1; ++j) {
        if (grid[i][j] == 1 && uf.find(i * n + j) != uf.find(m * n)) {
          ++ans;
        }
      }
    }
    return ans;
  }
}

class UnionFind {
public:
  UnionFind(int n) {
    p = vector<int>(n);
    size = vector<int>(n, 1);
    iota(p.begin(), p.end(), 0);
  }

  void unite(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa != pb) {
      if (size[pa] > size[pb]) {
        p[pb] = pa;
        size[pa] += size[pb];
      } else {
        p[pa] = pb;
        size[pb] += size[pa];
      }
    }
  }

  int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }

private:
  vector<int> p, size;
};

class Solution {
public:
  int numEnclaves(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    UnionFind uf(m * n + 1);
    int dirs[5] = {-1, 0, 1, 0, -1};
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 1) {
          if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
            uf.unite(i * n + j, m * n);
          } else {
            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) {
                uf.unite(i * n + j, x * n + y);
              }
            }
          }
        }
      }
    }
    int ans = 0;
    for (int i = 1; i < m - 1; ++i) {
      for (int j = 1; j < n - 1; ++j) {
        ans += grid[i][j] == 1 && uf.find(i * n + j) != uf.find(m * n);
      }
    }
    return ans;
  }
};

type unionFind struct {
  p, size []int
}

func newUnionFind(n int) *unionFind {
  p := make([]int, n)
  size := make([]int, n)
  for i := range p {
    p[i] = i
    size[i] = 1
  }
  return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
  if uf.p[x] != x {
    uf.p[x] = uf.find(uf.p[x])
  }
  return uf.p[x]
}

func (uf *unionFind) union(a, b int) {
  pa, pb := uf.find(a), uf.find(b)
  if pa != pb {
    if uf.size[pa] > uf.size[pb] {
      uf.p[pb] = pa
      uf.size[pa] += uf.size[pb]
    } else {
      uf.p[pa] = pb
      uf.size[pb] += uf.size[pa]
    }
  }
}

func numEnclaves(grid [][]int) (ans int) {
  m, n := len(grid), len(grid[0])
  uf := newUnionFind(m*n + 1)
  dirs := [5]int{-1, 0, 1, 0, -1}
  for i, row := range grid {
    for j, v := range row {
      if v == 1 {
        if i == 0 || i == m-1 || j == 0 || j == n-1 {
          uf.union(i*n+j, m*n)
        } else {
          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 {
              uf.union(i*n+j, x*n+y)
            }
          }
        }
      }
    }
  }
  for i, row := range grid {
    for j, v := range row {
      if v == 1 && uf.find(i*n+j) != uf.find(m*n) {
        ans++
      }
    }
  }
  return
}