1020.飞地的数量

链接：1020.飞地的数量
难度：Medium
标签：深度优先搜索、广度优先搜索、并查集、数组、矩阵
简介：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。

题解 1 - cpp

• 编辑时间：2022-02-12
• 执行用时：84ms
• 内存消耗：23.6MB
• 编程语言：cpp
• 解法介绍：uf,统计所有出口。

int dirs[4][2] = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
class UnionFind {
   public:
    vector<int> data;
    UnionFind(int size) : data(size) {
        for (int i = 0; i < size; i++) data[i] = i;
    }
    int find(int e) {
        if (data[e] == e) return e;
        return data[e] = find(data[e]);
    }
    void uni(int e1, int e2) { data[find(e2)] = find(e1); }
};
class Solution {
   public:
    int n, m;
    int get_idx(int row, int col) { return row * m + col; }
    int numEnclaves(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        int ans = 0;
        UnionFind uf(n * m);
        unordered_set<int> s1, s2;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 0) continue;
                ans++;
                if (i == 0 || j == 0 || i == n - 1 || j == m - 1)
                    s1.insert(get_idx(i, j));
                for (int k = 0; k < 4; k++) {
                    int ni = i + dirs[k][0], nj = j + dirs[k][1];
                    if (ni < 0 || ni >= n || nj < 0 || nj >= m ||
                        grid[ni][nj] == 0)
                        continue;
                    uf.uni(get_idx(i, j), get_idx(ni, nj));
                }
            }
        }
        for (auto& idx : s1) s2.insert(uf.find(idx));
        for (int i = 0; i < n * m; i++) {
            if (s2.count(uf.find(i))) ans--;
        }
        return ans;
    }
};

题解 2 - cpp

• 编辑时间：2022-02-12
• 执行用时：44ms
• 内存消耗：21.1MB
• 编程语言：cpp
• 解法介绍：dfs,对于每个出口进行遍历,遍历到的陆地都可出。

int dirs[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
class Solution {
   public:
    typedef pair<int, int> node;
    int m, n;
    int numEnclaves(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        for (int i = 0; i < n; i++) {
            if (grid[i][0]) {
                grid[i][0] = 0;
                dfs(grid, i, 0);
            }
            if (m > 1 && grid[i][m - 1]) {
                grid[i][m - 1] = 0;
                dfs(grid, i, m - 1);
            }
        }
        for (int i = 1; i < m - 1; i++) {
            if (grid[0][i]) {
                grid[0][i] = 0;
                dfs(grid, 0, i);
            }
            if (n > 1 && grid[n - 1][i]) {
                grid[n - 1][i] = 0;
                dfs(grid, n - 1, i);
            }
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j]) ans++;
            }
        }
        return ans;
    }
    void dfs(vector<vector<int>>& grid, const int row, const int col) {
        for (int i = 0; i < 4; i++) {
            int nrow = row + dirs[i][0], ncol = col + dirs[i][1];
            if (nrow < 0 || nrow >= n || ncol < 0 || ncol >= m ||
                grid[nrow][ncol] == 0)
                continue;
            grid[nrow][ncol] = 0;
            dfs(grid, nrow, ncol);
        }
    }
};

题解 3 - cpp

• 编辑时间：2022-02-12
• 执行用时：48ms
• 内存消耗：21.6MB
• 编程语言：cpp
• 解法介绍：bfs,对于每个出口进行遍历,遍历到的陆地都可出。

int dirs[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
class Solution {
   public:
    typedef pair<int, int> node;
    int m, n;
    int numEnclaves(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        queue<node> q;
        for (int i = 0; i < n; i++) {
            if (grid[i][0]) {
                q.push(make_pair(i, 0));
                grid[i][0] = 0;
            }
            if (m > 1 && grid[i][m - 1]) {
                q.push(make_pair(i, m - 1));
                grid[i][m - 1] = 0;
            }
        }
        for (int i = 1; i < m - 1; i++) {
            if (grid[0][i]) {
                q.push(make_pair(0, i));
                grid[0][i] = 0;
            }
            if (n > 1 && grid[n - 1][i]) {
                q.push(make_pair(n - 1, i));
                grid[n - 1][i] = 0;
            }
        }
        while (q.size()) {
            node v = q.front();
            q.pop();
            for (int i = 0; i < 4; i++) {
                int nrow = v.first + dirs[i][0], ncol = v.second + dirs[i][1];
                if (nrow < 0 || nrow >= n || ncol < 0 || ncol >= m ||
                    grid[nrow][ncol] == 0)
                    continue;
                q.push(make_pair(nrow, ncol));
                grid[nrow][ncol] = 0;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j]) ans++;
            }
        }
        return ans;
    }
};