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.lengthn == grid[i].length1 <= m, n <= 500grid[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。
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16 | 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)
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36 | 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);
}
}
}
}
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30 | 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;
}
};
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27 | 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
}
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29 | 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;
}
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37 | 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
}
}
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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。
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18 | 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)
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33 | 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;
}
}
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34 | 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;
}
};
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30 | 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
}
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32 | 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;
}
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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。
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41 | 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)
)
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67 | 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;
}
}
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63 | 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;
}
};
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63 | 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
}
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