LeetCode 130. Surrounded Regions (并查集)

Tags:
- Breadth-first Search
- Union Find

问题描述

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 XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

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解题思路

题意是把所有不与边界直接相连的O变为X。
使用并查集，将所有相连的O转换为并查集并且相连。
然后定义一个虚节点，表示边界。
把与边界相连的结点和虚节点链接，表示该节点与边界相连，不需要变成X。
最后把不与虚节点连接的O结点变为X。

参考代码

#include <iostream>#include <vector>using namespace std;class UF                // union find{private:    int *id;            // parent link    int *sz;            // size of component for roots    int count;          // number of componentspublic:    UF(int N)    {        count = N;        id = new int[N];        sz = new int[N];        for (int i = 0; i < N; ++i)        {            id[i] = i;            sz[i] = 1;        }    }    ~UF()    {        delete[] id;        delete[] sz;    }    int getCount()    {        return count;    }    int find(int i)    {        while (i != id[i])        {            id[i] = id[id[i]];                  // path compression by halving            i = id[i];        }        return i;    }    bool connected(int p, int q)                // check if two union is connected    {        return find(p) == find(q);    }    void connect(int p, int q)                  // merge    {        int i = find(p);        int j = find(q);        if (i == j) return;        // make smaller root point to larger one        if(sz[i] < sz[j])        {            id[i] = j;            sz[j] += sz[i];        }        else        {            id[j] = i;            sz[i] += sz[j];        }        count--;    }};class Solution{public:    void solve(vector<vector<char>> &board)    {        int n = board.size();        if (n == 0)return;        int m = board[0].size();        UF uf = UF(n * m + 1);          // n * m is a dummy node, which means boundary        for (int i = 0; i < n; ++i)        {            for (int j = 0; j < m; ++j)            {                // if a 'O' node is on the boundary, connect it to the dummy node                if((i == 0 || i == n - 1 || j == 0 || j == m - 1) && board[i][j] == 'O')                {                    uf.connect(i * m + j, n * m);                }                // connect a 'O' node to its neighbor 'O' nodes                else if (board[i][j] == 'O')                {                    if (board[i - 1][j] == 'O')                        uf.connect(i * m + j, (i - 1) * m + j);                    if (board[i + 1][j] == 'O')                        uf.connect(i * m + j, (i + 1) * m + j);                    if (board[i][j - 1] == 'O')                        uf.connect(i * m + j, i * m + j - 1);                    if (board[i][j + 1] == 'O')                        uf.connect(i * m + j, i * m + j + 1);                }            }        }        for (int i = 0; i < n; ++i)        {            for (int j = 0; j < m; ++j)            {                // if a 'O' node is not connected to the dummy node, it is captured                if (!uf.connected(i * m + j, n * m))                {                    board[i][j] = 'X';                }            }        }    }};int main(){    vector<vector<char>> board =    {        {'X', 'X', 'X', 'X'},        {'X', 'O', 'O', 'X'} ,        {'X', 'X', 'O', 'X'},        {'X', 'O', 'X', 'X'}    };    auto sl = new Solution();    sl->solve(board);    for (auto a : board)    {        for (auto b : a)        {            cout << b << " ";        }        cout << endl;    }    system("pause");    return 0;}