Question

1254. 统计封闭岛屿的数目

English Version

题目描述

二维矩阵 grid 由 0 （土地）和 1 （水）组成。岛是由最大的4个方向连通的 0 组成的群，封闭岛是一个 完全 由1包围（左、上、右、下）的岛。

请返回 _封闭岛屿_ 的数目。

 

示例 1：

输入：grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
输出：2
解释：
灰色区域的岛屿是封闭岛屿，因为这座岛屿完全被水域包围（即被 1 区域包围）。

示例 2：

输入：grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
输出：1

示例 3：

输入：grid = [[1,1,1,1,1,1,1],
       [1,0,0,0,0,0,1],
       [1,0,1,1,1,0,1],
       [1,0,1,0,1,0,1],
       [1,0,1,1,1,0,1],
       [1,0,0,0,0,0,1],
       [1,1,1,1,1,1,1]]
输出：2

 

提示：

• 1 <= grid.length, grid[0].length <= 100
• 0 <= grid[i][j] <=1

解法

方法一：DFS

遍历矩阵，对于每个陆地，我们进行深度优先搜索，找到与其相连的所有陆地，然后判断是否存在边界上的陆地，如果存在，则不是封闭岛屿，否则是封闭岛屿，答案加一。

最后返回答案即可。

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

class Solution:
  def closedIsland(self, grid: List[List[int]]) -> int:
    def dfs(i: int, j: int) -> int:
      res = int(0 < i < m - 1 and 0 < j < n - 1)
      grid[i][j] = 1
      for a, b in pairwise(dirs):
        x, y = i + a, j + b
        if 0 <= x < m and 0 <= y < n and grid[x][y] == 0:
          res &= dfs(x, y)
      return res

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

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

  public int closedIsland(int[][] grid) {
    m = grid.length;
    n = grid[0].length;
    this.grid = grid;
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 0) {
          ans += dfs(i, j);
        }
      }
    }
    return ans;
  }

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

class Solution {
public:
  int closedIsland(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    int ans = 0;
    int dirs[5] = {-1, 0, 1, 0, -1};
    function<int(int, int)> dfs = [&](int i, int j) -> int {
      int res = i > 0 && i < m - 1 && j > 0 && j < n - 1 ? 1 : 0;
      grid[i][j] = 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] == 0) {
          res &= dfs(x, y);
        }
      }
      return res;
    };
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans += grid[i][j] == 0 && dfs(i, j);
      }
    }
    return ans;
  }
};

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

function closedIsland(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): number => {
    let res = i > 0 && j > 0 && i < m - 1 && j < n - 1 ? 1 : 0;
    grid[i][j] = 1;
    for (let k = 0; k < 4; ++k) {
      const [x, y] = [i + dirs[k], j + dirs[k + 1]];
      if (x >= 0 && y >= 0 && x < m && y < n && grid[x][y] === 0) {
        res &= dfs(x, y);
      }
    }
    return res;
  };
  let ans = 0;
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; j++) {
      if (grid[i][j] === 0) {
        ans += dfs(i, j);
      }
    }
  }
  return ans;
}

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

  public int ClosedIsland(int[][] grid) {
    m = grid.Length;
    n = grid[0].Length;
    this.grid = grid;
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 0) {
          ans += dfs(i, j);
        }
      }
    }
    return ans;
  }

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

方法二：并查集

我们可以用并查集维护每一块连通的陆地。

遍历矩阵，如果当前位置是在边界上，我们将其与虚拟节点 $m \times n$ 连接。如果当前位置是陆地，我们将其与下方和右方的陆地连接。

接着，我们再次遍历矩阵，对于每一块陆地，如果其根节点就是本身，那么答案加一。

最后返回答案即可。

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

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

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

  def union(self, a: int, b: int):
    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 closedIsland(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    uf = UnionFind(m * n + 1)
    for i in range(m):
      for j in range(n):
        if i == 0 or i == m - 1 or j == 0 or j == n - 1:
          uf.union(i * n + j, m * n)
        if grid[i][j] == 0:
          if i < m - 1 and grid[i + 1][j] == 0:
            uf.union(i * n + j, (i + 1) * n + j)
          if j < n - 1 and grid[i][j + 1] == 0:
            uf.union(i * n + j, i * n + j + 1)
    ans = 0
    for i in range(m):
      for j in range(n):
        ans += grid[i][j] == 0 and uf.find(i * n + j) == i * n + j
    return ans

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 closedIsland(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    UnionFind uf = new UnionFind(m * n + 1);
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
          uf.union(i * n + j, m * n);
        }
        if (grid[i][j] == 0) {
          if (i + 1 < m && grid[i + 1][j] == 0) {
            uf.union(i * n + j, (i + 1) * n + j);
          }
          if (j + 1 < n && grid[i][j + 1] == 0) {
            uf.union(i * n + j, i * n + j + 1);
          }
        }
      }
    }
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 0 && uf.find(i * n + j) == i * n + j) {
          ++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 closedIsland(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    UnionFind uf(m * n + 1);
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
          uf.unite(i * n + j, m * n);
        }
        if (grid[i][j] == 0) {
          if (i + 1 < m && grid[i + 1][j] == 0) {
            uf.unite(i * n + j, (i + 1) * n + j);
          }
          if (j + 1 < n && grid[i][j + 1] == 0) {
            uf.unite(i * n + j, i * n + j + 1);
          }
        }
      }
    }
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans += grid[i][j] == 0 && uf.find(i * n + j) == i * n + j;
      }
    }
    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 closedIsland(grid [][]int) (ans int) {
  m, n := len(grid), len(grid[0])
  uf := newUnionFind(m*n + 1)
  for i, row := range grid {
    for j, v := range row {
      if i == 0 || i == m-1 || j == 0 || j == n-1 {
        uf.union(i*n+j, m*n)
      }
      if v == 0 {
        if i+1 < m && grid[i+1][j] == 0 {
          uf.union(i*n+j, (i+1)*n+j)
        }
        if j+1 < n && grid[i][j+1] == 0 {
          uf.union(i*n+j, i*n+j+1)
        }
      }
    }
  }
  for i, row := range grid {
    for j, v := range row {
      if v == 0 && uf.find(i*n+j) == i*n+j {
        ans++
      }
    }
  }
  return
}

function closedIsland(grid: number[][]): number {
  const m = grid.length;
  const n = grid[0].length;
  const uf = new UnionFind(m * n + 1);
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      if (i === 0 || i === m - 1 || j === 0 || j === n - 1) {
        uf.union(i * n + j, m * n);
      }
      if (grid[i][j] === 0) {
        if (i + 1 < m && grid[i + 1][j] === 0) {
          uf.union(i * n + j, (i + 1) * n + j);
        }
        if (j + 1 < n && grid[i][j + 1] === 0) {
          uf.union(i * n + j, i * n + j + 1);
        }
      }
    }
  }
  let ans = 0;
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; j++) {
      if (grid[i][j] === 0 && uf.find(i * n + j) === i * n + j) {
        ++ans;
      }
    }
  }
  return ans;
}

class UnionFind {
  private p: number[];
  private size: number[];

  constructor(n: number) {
    this.p = Array(n)
      .fill(0)
      .map((_, i) => i);
    this.size = Array(n).fill(1);
  }

  find(x: number): number {
    if (this.p[x] !== x) {
      this.p[x] = this.find(this.p[x]);
    }
    return this.p[x];
  }

  union(a: number, b: number): void {
    const [pa, pb] = [this.find(a), this.find(b)];
    if (pa === pb) {
      return;
    }
    if (this.size[pa] > this.size[pb]) {
      this.p[pb] = pa;
      this.size[pa] += this.size[pb];
    } else {
      this.p[pa] = pb;
      this.size[pb] += this.size[pa];
    }
  }
}

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];
      }
    }
  }
}

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