Question

827. 最大人工岛

English Version

题目描述

给你一个大小为 n x n 二进制矩阵 grid 。最多 只能将一格 0 变成 1 。

返回执行此操作后，grid 中最大的岛屿面积是多少？

岛屿 由一组上、下、左、右四个方向相连的 1 形成。

 

示例 1:

输入: grid = [[1, 0], [0, 1]]
输出: 3
解释: 将一格0变成1，最终连通两个小岛得到面积为 3 的岛屿。

示例 2:

输入: grid = [[1, 1], [1, 0]]
输出: 4
解释: 将一格0变成1，岛屿的面积扩大为 4。

示例 3:

输入: grid = [[1, 1], [1, 1]]
输出: 4
解释: 没有0可以让我们变成1，面积依然为 4。

 

提示：

• n == grid.length
• n == grid[i].length
• 1 <= n <= 500
• grid[i][j] 为 0 或 1

解法

方法一：并查集

并查集是一种树形的数据结构，顾名思义，它用于处理一些不交集的合并及查询问题。 它支持两种操作：

1. 查找（Find）：确定某个元素处于哪个子集，单次操作时间复杂度 $O(\alpha(n))$
2. 合并（Union）：将两个子集合并成一个集合，单次操作时间复杂度 $O(\alpha(n))$

其中 $\alpha$ 为阿克曼函数的反函数，其增长极其缓慢，也就是说其单次操作的平均运行时间可以认为是一个很小的常数。

以下是并查集的常用模板，需要熟练掌握。其中：

• n 表示节点数
• p 存储每个点的父节点，初始时每个点的父节点都是自己
• size 只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
• find(x) 函数用于查找 $x$ 所在集合的祖宗节点
• union(a, b) 函数用于合并 $a$ 和 $b$ 所在的集合

p = list(range(n))
size = [1] * n


def find(x):
  if p[x] != x:
    # 路径压缩
    p[x] = find(p[x])
  return p[x]


def union(a, b):
  pa, pb = find(a), find(b)
  if pa == pb:
    return
  p[pa] = pb
  size[pb] += size[pa]

在这道题中，相邻的 $1$ 组成一个岛屿，因此，我们需要将相邻的 $1$ 归到同一个集合中。这可以视为一个合并操作，不难想到用并查集来实现。

第一次遍历 grid，通过并查集的 union 操作合并所有相邻的 $1$，并且统计每个岛屿的面积，记录在 $size$ 数组中。

再次遍历 grid，对于每个 $0$，我们统计相邻的四个点中 $1$ 所在的岛屿（通过并查集的 find 操作找到所在岛屿），累加去重后的岛屿面积，更新最大值。

时间复杂度 $O(n^2\times \alpha(n))$。其中 $n$ 为矩阵 grid 的边长。

class Solution:
  def largestIsland(self, grid: List[List[int]]) -> int:
    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    def union(a, b):
      pa, pb = find(a), find(b)
      if pa == pb:
        return
      p[pa] = pb
      size[pb] += size[pa]

    n = len(grid)
    p = list(range(n * n))
    size = [1] * (n * n)
    for i, row in enumerate(grid):
      for j, v in enumerate(row):
        if v:
          for a, b in [[0, -1], [-1, 0]]:
            x, y = i + a, j + b
            if 0 <= x < n and 0 <= y < n and grid[x][y]:
              union(x * n + y, i * n + j)
    ans = max(size)
    for i, row in enumerate(grid):
      for j, v in enumerate(row):
        if v == 0:
          vis = set()
          t = 1
          for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
            x, y = i + a, j + b
            if 0 <= x < n and 0 <= y < n and grid[x][y]:
              root = find(x * n + y)
              if root not in vis:
                vis.add(root)
                t += size[root]
          ans = max(ans, t)
    return ans

class Solution {
  private int n;
  private int[] p;
  private int[] size;
  private int ans = 1;
  private int[] dirs = new int[] {-1, 0, 1, 0, -1};

  public int largestIsland(int[][] grid) {
    n = grid.length;
    p = new int[n * n];
    size = new int[n * n];
    for (int i = 0; i < p.length; ++i) {
      p[i] = i;
      size[i] = 1;
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (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 < n && y >= 0 && y < n && grid[x][y] == 1) {
              int pa = find(x * n + y), pb = find(i * n + j);
              if (pa == pb) {
                continue;
              }
              p[pa] = pb;
              size[pb] += size[pa];
              ans = Math.max(ans, size[pb]);
            }
          }
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 0) {
          int t = 1;
          Set<Integer> vis = new HashSet<>();
          for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1) {
              int root = find(x * n + y);
              if (!vis.contains(root)) {
                vis.add(root);
                t += size[root];
              }
            }
          }
          ans = Math.max(ans, t);
        }
      }
    }
    return ans;
  }

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

class Solution {
public:
  const static inline vector<int> dirs = {-1, 0, 1, 0, -1};

  int largestIsland(vector<vector<int>>& grid) {
    int n = grid.size();
    vector<int> p(n * n);
    vector<int> size(n * n, 1);
    iota(p.begin(), p.end(), 0);

    function<int(int)> find;
    find = [&](int x) {
      if (p[x] != x) {
        p[x] = find(p[x]);
      }
      return p[x];
    };

    int ans = 1;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j]) {
          for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y]) {
              int pa = find(x * n + y), pb = find(i * n + j);
              if (pa == pb) continue;
              p[pa] = pb;
              size[pb] += size[pa];
              ans = max(ans, size[pb]);
            }
          }
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (!grid[i][j]) {
          int t = 1;
          unordered_set<int> vis;
          for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y]) {
              int root = find(x * n + y);
              if (!vis.count(root)) {
                vis.insert(root);
                t += size[root];
              }
            }
          }
          ans = max(ans, t);
        }
      }
    }
    return ans;
  }
};

func largestIsland(grid [][]int) int {
  n := len(grid)
  p := make([]int, n*n)
  size := make([]int, n*n)
  for i := range p {
    p[i] = i
    size[i] = 1
  }
  var find func(int) int
  find = func(x int) int {
    if p[x] != x {
      p[x] = find(p[x])
    }
    return p[x]
  }
  dirs := []int{-1, 0, 1, 0, -1}
  ans := 1
  for i, row := range grid {
    for j, v := range row {
      if v == 1 {
        for k := 0; k < 4; k++ {
          x, y := i+dirs[k], j+dirs[k+1]
          if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 {
            pa, pb := find(x*n+y), find(i*n+j)
            if pa != pb {
              p[pa] = pb
              size[pb] += size[pa]
              ans = max(ans, size[pb])
            }
          }
        }
      }
    }
  }
  for i, row := range grid {
    for j, v := range row {
      if v == 0 {
        t := 1
        vis := map[int]struct{}{}
        for k := 0; k < 4; k++ {
          x, y := i+dirs[k], j+dirs[k+1]
          if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 {
            root := find(x*n + y)
            if _, ok := vis[root]; !ok {
              vis[root] = struct{}{}
              t += size[root]
            }
          }
        }
        ans = max(ans, t)
      }
    }
  }
  return ans
}

function largestIsland(grid: number[][]): number {
  const n = grid.length;
  const vis = Array.from({ length: n }, () => new Array(n).fill(false));
  const group = Array.from({ length: n }, () => new Array(n).fill(0));
  const dfs = (i: number, j: number, paths: [number, number][]) => {
    if (i < 0 || j < 0 || i === n || j === n || vis[i][j] || grid[i][j] !== 1) {
      return;
    }
    vis[i][j] = true;
    paths.push([i, j]);
    dfs(i + 1, j, paths);
    dfs(i, j + 1, paths);
    dfs(i - 1, j, paths);
    dfs(i, j - 1, paths);
  };
  let count = 1;
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      const paths: [number, number][] = [];
      dfs(i, j, paths);
      if (paths.length !== 0) {
        for (const [x, y] of paths) {
          group[x][y] = count;
          grid[x][y] = paths.length;
        }
        count++;
      }
    }
  }

  let res = 0;
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      let sum = grid[i][j];
      if (grid[i][j] === 0) {
        sum++;
        const set = new Set();
        if (i !== 0) {
          sum += grid[i - 1][j];
          set.add(group[i - 1][j]);
        }
        if (i !== n - 1 && !set.has(group[i + 1][j])) {
          sum += grid[i + 1][j];
          set.add(group[i + 1][j]);
        }
        if (j !== 0 && !set.has(group[i][j - 1])) {
          sum += grid[i][j - 1];
          set.add(group[i][j - 1]);
        }
        if (j !== n - 1 && !set.has(group[i][j + 1])) {
          sum += grid[i][j + 1];
        }
      }
      res = Math.max(res, sum);
    }
  }
  return res;
}

use std::collections::HashSet;
impl Solution {
  fn dfs(
    i: usize,
    j: usize,
    grid: &Vec<Vec<i32>>,
    paths: &mut Vec<(usize, usize)>,
    vis: &mut Vec<Vec<bool>>
  ) {
    let n = vis.len();
    if vis[i][j] || grid[i][j] != 1 {
      return;
    }
    paths.push((i, j));
    vis[i][j] = true;
    if i != 0 {
      Self::dfs(i - 1, j, grid, paths, vis);
    }
    if j != 0 {
      Self::dfs(i, j - 1, grid, paths, vis);
    }
    if i != n - 1 {
      Self::dfs(i + 1, j, grid, paths, vis);
    }
    if j != n - 1 {
      Self::dfs(i, j + 1, grid, paths, vis);
    }
  }

  pub fn largest_island(mut grid: Vec<Vec<i32>>) -> i32 {
    let n = grid.len();
    let mut vis = vec![vec![false; n]; n];
    let mut group = vec![vec![0; n]; n];
    let mut count = 1;
    for i in 0..n {
      for j in 0..n {
        let mut paths: Vec<(usize, usize)> = Vec::new();
        Self::dfs(i, j, &grid, &mut paths, &mut vis);
        let m = paths.len() as i32;
        if m != 0 {
          for (x, y) in paths {
            grid[x][y] = m;
            group[x][y] = count;
          }
          count += 1;
        }
      }
    }
    let mut res = 0;
    for i in 0..n {
      for j in 0..n {
        let mut sum = grid[i][j];
        if grid[i][j] == 0 {
          sum += 1;
          let mut set = HashSet::new();
          if i != 0 {
            sum += grid[i - 1][j];
            set.insert(group[i - 1][j]);
          }
          if j != 0 && !set.contains(&group[i][j - 1]) {
            sum += grid[i][j - 1];
            set.insert(group[i][j - 1]);
          }
          if i != n - 1 && !set.contains(&group[i + 1][j]) {
            sum += grid[i + 1][j];
            set.insert(group[i + 1][j]);
          }
          if j != n - 1 && !set.contains(&group[i][j + 1]) {
            sum += grid[i][j + 1];
            set.insert(group[i][j + 1]);
          }
        }
        res = res.max(sum);
      }
    }
    res
  }
}

方法二：DFS

我们也可以通过 DFS，找到每个岛屿。

同一个岛屿中的所有点都属于同一个集合，我们可以用不同的 root 值标识不同的岛屿，用 $p$ 记录每个 $grid[i][j]$ 对应的 root 值，用 $cnt$ 记录每个岛屿的面积。

遍历 grid，对于每个 $0$，我们统计相邻的四个点中 $1$ 所在的岛屿（与方法一不同的是，我们这里直接取 $p[i][j]$ 作为 root），累加去重后的岛屿面积，更新最大值。

时间复杂度 $O(n^2)$。其中 $n$ 为矩阵 grid 的边长。

class Solution:
  def largestIsland(self, grid: List[List[int]]) -> int:
    def dfs(i, j):
      p[i][j] = root
      cnt[root] += 1
      for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
        x, y = i + a, j + b
        if 0 <= x < n and 0 <= y < n and grid[x][y] and p[x][y] == 0:
          dfs(x, y)

    n = len(grid)
    cnt = Counter()
    p = [[0] * n for _ in range(n)]
    root = 0
    for i, row in enumerate(grid):
      for j, v in enumerate(row):
        if v and p[i][j] == 0:
          root += 1
          dfs(i, j)

    ans = max(cnt.values(), default=0)
    for i, row in enumerate(grid):
      for j, v in enumerate(row):
        if v == 0:
          t = 1
          vis = set()
          for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
            x, y = i + a, j + b
            if 0 <= x < n and 0 <= y < n:
              root = p[x][y]
              if root not in vis:
                vis.add(root)
                t += cnt[root]
          ans = max(ans, t)
    return ans

class Solution {
  private int n;
  private int ans;
  private int root;
  private int[][] p;
  private int[][] grid;
  private int[] cnt;
  private int[] dirs = new int[] {-1, 0, 1, 0, -1};

  public int largestIsland(int[][] grid) {
    n = grid.length;
    cnt = new int[n * n + 1];
    p = new int[n][n];
    this.grid = grid;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 1 && p[i][j] == 0) {
          ++root;
          dfs(i, j);
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == 0) {
          int t = 1;
          Set<Integer> vis = new HashSet<>();
          for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < n && y >= 0 && y < n) {
              int root = p[x][y];
              if (!vis.contains(root)) {
                vis.add(root);
                t += cnt[root];
              }
            }
          }
          ans = Math.max(ans, t);
        }
      }
    }
    return ans;
  }

  private void dfs(int i, int j) {
    p[i][j] = root;
    ++cnt[root];
    ans = Math.max(ans, cnt[root]);
    for (int k = 0; k < 4; ++k) {
      int x = i + dirs[k], y = j + dirs[k + 1];
      if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 && p[x][y] == 0) {
        dfs(x, y);
      }
    }
  }
}

class Solution {
public:
  const static inline vector<int> dirs = {-1, 0, 1, 0, -1};

  int largestIsland(vector<vector<int>>& grid) {
    int n = grid.size();
    int ans = 0;
    int root = 0;
    vector<vector<int>> p(n, vector<int>(n));
    vector<int> cnt(n * n + 1);

    function<void(int, int)> dfs;
    dfs = [&](int i, int j) {
      p[i][j] = root;
      ++cnt[root];
      ans = max(ans, cnt[root]);
      for (int k = 0; k < 4; ++k) {
        int x = i + dirs[k], y = j + dirs[k + 1];
        if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] && p[x][y] == 0) {
          dfs(x, y);
        }
      }
    };

    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] && p[i][j] == 0) {
          ++root;
          dfs(i, j);
        }
      }
    }

    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (!grid[i][j]) {
          int t = 1;
          unordered_set<int> vis;
          for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < n && y >= 0 && y < n) {
              int root = p[x][y];
              if (!vis.count(root)) {
                vis.insert(root);
                t += cnt[root];
              }
            }
          }
          ans = max(ans, t);
        }
      }
    }
    return ans;
  }
};

func largestIsland(grid [][]int) int {
  n := len(grid)
  p := make([][]int, n)
  for i := range p {
    p[i] = make([]int, n)
  }
  cnt := make([]int, n*n+1)
  dirs := []int{-1, 0, 1, 0, -1}
  ans, root := 0, 0

  var dfs func(i, j int)
  dfs = func(i, j int) {
    p[i][j] = root
    cnt[root]++
    ans = max(ans, cnt[root])
    for k := 0; k < 4; k++ {
      x, y := i+dirs[k], j+dirs[k+1]
      if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 && p[x][y] == 0 {
        dfs(x, y)
      }
    }
  }

  for i, row := range grid {
    for j, v := range row {
      if v == 1 && p[i][j] == 0 {
        root++
        dfs(i, j)
      }
    }
  }
  for i, row := range grid {
    for j, v := range row {
      if v == 0 {
        t := 1
        vis := map[int]struct{}{}
        for k := 0; k < 4; k++ {
          x, y := i+dirs[k], j+dirs[k+1]
          if x >= 0 && x < n && y >= 0 && y < n {
            root := p[x][y]
            if _, ok := vis[root]; !ok {
              vis[root] = struct{}{}
              t += cnt[root]
            }
          }
        }
        ans = max(ans, t)
      }
    }
  }
  return ans
}