200. Number of Islands+130. Surrounded Regions(並查集/DFS)

FreeeLinux發表於2017-02-10
並查集

首先看200題：

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000
Answer: 1

Example 2:

11000
11000
00100
00011

DFS：

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        if(grid.empty())
            return 0;
        m = grid.size();
        n = grid[0].size();
        int counter = 0;
        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(grid[i][j] == '1'){
                    dfs_mark(grid, i, j);
                    ++counter;
                }
            }
        }
        return counter;
    }
    void dfs_mark(vector<vector<char>>& grid, int x, int y){
        if(x < 0 || x >= m || y < 0 || y >= n || grid[x][y] != '1')
            return;
        grid[x][y] = '0';
        dfs_mark(grid, x-1, y);
        dfs_mark(grid, x+1, y);
        dfs_mark(grid, x, y-1);
        dfs_mark(grid, x, y+1);
    }
private:
    int m;
    int n;
};

並查集版本：

class Solution {
public:
    class uf_set {
    public:
        uf_set(const vector<vector<char>>& grid, int m, int n) 
            : father_(new int[m*n]), rnk_(new int[m*n]), counter_(0){
                fill(rnk_, rnk_+m*n, 0);
                for(int i=0; i<m; ++i){
                    for(int j=0; j<n; ++j){
                        if(grid[i][j] == '1'){
                            int id = i * n + j;
                            father_[id] = id;
                            ++counter_;
                        }
                    }
                }
        }
        ~uf_set(){
            delete []father_;
            delete []rnk_;
        }
    public:
        int find(int x){
            int root = x;
            while(root != father_[root])
                root = father_[root];

            while(x != father_[x]){
                int y = father_[x];
                father_[x] = root;
                x = y;
            }   
            return root;
        }
        void un(int x, int y){
            x = find(x), y = find(y);
            if(x == y) return ;
            if(rnk_[x] > rnk_[y]) father_[y] = x;
            else father_[x] = y, rnk_[y] += rnk_[x] == rnk_[y];
            --counter_;
        }
        int counter() const { return counter_; }

    private:
        int* father_;
        int* rnk_;
        int  counter_;
    };
public:
    vector<vector<char>> direction = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    int numIslands(vector<vector<char>>& grid) {
        if(grid.empty())
            return 0;
        int m = grid.size(), n = grid[0].size();
        uf_set us(grid, m, n);
        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(grid[i][j] == '1'){
                    for(auto d: direction){
                        int x = i + d[0];
                        int y = j + d[1];
                        if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1'){
                            int id1 = i * n + j;
                            int id2 = x * n + y;
                            us.un(id1, id2);
                        }
                    }
                }
            }
        }
        int res = us.counter();
        return res;
    }
};

做上面這道題時，我強迫症發作，所以每個成員變數後面都加了字尾。後了一想，這樣寫專案很有用，寫演算法就是反而很雞肋。於是我決定，以後寫演算法不在成員變數後面加字尾下劃線了。

看130題。

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X
Subscribe to see which companies asked this question.

首先我直接用DFS：

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        if(board.empty())
            return ;
        int m = board.size();
        int n = board[0].size();
        for(int i=0; i<m; ++i)
            if(board[i][0] == 'O') dfs_mark(board, m, n, i, 0);
        for(int j=1; j<n; ++j)
            if(board[0][j] == 'O') dfs_mark(board, m, n, 0, j);    
        for(int i=1; i<m; ++i)
            if(board[i][n-1] == 'O') dfs_mark(board, m, n, i, n-1);
        for(int j=1; j<n-1; ++j)
            if(board[m-1][j] == 'O') dfs_mark(board, m, n, m-1, j);
        for(int i=1; i<m-1; ++i){
            for(int j=1; j<n-1; ++j)
                if(board[i][j] != 'N') board[i][j] = 'X';
        }
        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j)
                if(board[i][j] == 'N') board[i][j] = 'O';
        }
    }
    void dfs_mark(vector<vector<char>>& board, int m, int n, int x, int y){
        if(x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'O') return;
        board[x][y] = 'N';
        dfs_mark(board, m, n, x-1, y);
        dfs_mark(board, m, n, x+1, y);
        dfs_mark(board, m, n, x, y-1);
        dfs_mark(board, m, n, x, y+1);
    }
};

然後超時了。但是我在網上搜這道題，在別的OJ上用這份程式碼通過了。所以leetcode要求還是蠻嚴格的。

然後是並查集版本：

class Solution {
public:
    vector<vector<char>> direction = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    void solve(vector<vector<char>>& board) {
        if(board.empty())
            return;
        int m = board.size();
        int n = board[0].size();
        uf_set us(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) && board[i][j] == 'O')
                    us.union_set(i*n+j, m*n);
                else if(board[i][j] == 'O'){
                    for(auto d : direction){
                        int x = i + d[0];
                        int y = j + d[1];
                        if(board[x][y] == 'O')
                            us.union_set(i*n+j, x*n+y);
                    }
                }
            }
        }

        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(us.find_set(i*n+j) != us.find_set(n*m))
                    board[i][j] = 'X';
            }
        }
    }
private:
    class uf_set {
    public:
        uf_set(int sz) : father(new int[sz]), rank(new int[sz]) {
            for(int i=0; i<sz; ++i){
                father[i] = i;
                rank[i] = 0;
            }
        }
        int find_set(int x){
            int root = x;
            while(root != father[root])
                root = father[root];

            while(x != father[x]){
                int y = father[x];
                father[x] = root;
                x = y;
            }
            return root;
        }
        void union_set(int x, int y){
            x = find_set(x), y = find_set(y);
            if(x == y) return;
            if(rank[x] > rank[y]) 
                father[y] = x;
            else 
                father[x] = y, rank[y] += rank[x] == rank[y];
        }
    private:
        int* father;
        int* rank;
    };
};

秒殺之。

相關文章