130. Surrounded Regions 将包围的符号变换 BFS & DFS & UNION find

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

1.DFS

class Solution {public:    //1.用深度搜索的方法    void mark(vector<vector<char>>& board, int i, int j, int m, int n){        if(i > 1 && board[i-1][j] == 'O'){            board[i-1][j] = '1';            mark(board, i-1, j, m, n);        }        if(i+1 < m && board[i+1][j] == 'O'){            board[i+1][j] = '1';            mark(board, i+1, j, m, n);        }        if(j > 1 && board[i][j-1] == 'O'){            board[i][j-1] = '1';            mark(board, i, j-1, m, n);        }        if(j+1 < n && board[i][j+1] == 'O'){            board[i][j+1] = '1';            mark(board, i, j+1, m, n);        }            }        void solve(vector<vector<char>>& board) {        int m = board.size();        if(m == 0) return;        int n = board[0].size();        if(n == 0) return;        for(int i = 0; i < n; i++){ //对头尾两行进行边缘‘O’的查找            if(board[0][i] == 'O'){                board[0][i] = '1';                mark(board, 0, i, m, n);            }            if(m > 1){                if(board[m-1][i] == 'O'){                    board[m-1][i] = '1';                    mark(board, m-1, i, m, n);                }            }        }                for(int j = 0; j < m; j++){//对头尾两列进行边缘‘O’的查找            if(board[j][0] == 'O'){                board[j][0] = '1';                mark(board, j, 0, m, n);            }            if(n > 1){                if(board[j][n-1] == 'O'){                    board[j][n-1] = '1';                    mark(board, j, n-1, m, n);                }            }        }                for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(board[i][j] == 'O')                    board[i][j] = 'X';            }        }                for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(board[i][j] == '1')                    board[i][j] = 'O';            }        }    }};

2.BFS

class Solution {public:    void bfs(vector<vector<char>>& board, int x, int y, int m, int n){        queue<pair<int, int>>q;        q.push(make_pair(x,y));        board[x][y] = '+';                while(!q.empty()){            pair<int, int>cur = q.front();            q.pop();            pair<int, int>pos[4] = {{cur.first-1, cur.second}, {cur.first+1, cur.second},{cur.first, cur.second+1}, {cur.first, cur.second-1}};            for(int i = 0;i < 4; i++){                int xx = pos[i].first;                int yy = pos[i].second;                if(xx >= 0 && xx < m && yy >= 0 && yy < n && board[xx][yy] == 'O'){                    q.push(pos[i]);                    board[xx][yy] = '+';                }            }        }    }        void solve(vector<vector<char>>& board) {        //BFS        int m = board.size();        if(m == 0) return;        int n = board[0].size();        if(n == 0) return;        for(int i = 0; i < n; i++){            if(board[0][i] == 'O')                bfs(board, 0, i, m, n);            if(board[m-1][i] == 'O')                bfs(board, m-1, i, m, n);        }        for(int i = 0; i < m; i++){            if(board[i][0] == 'O')                bfs(board, i, 0, m, n);            if(board[i][n-1] == 'O')                bfs(board, i, n-1, m, n);        }        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(board[i][j] == 'O')                board[i][j] = 'X';            }        }                for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(board[i][j] == '+')                    board[i][j] = 'O';            }        }    }};

3.union find

别人的思想；

并查集常用来解决连通性的问题，即将一个图中连通的部分划分出来。当我们判断图中两个点之间是否存在路径时，就可以根据判断他们是否在一个连通区域。 
当数据较多时，并查集比BFS和DFS算法适用。 
但是，并查集的缺陷就是不能够求出连通的两个点之间的路径，但是DFS算法或者BFS算法适用。 
并查集的思想就是，同一个连通区域内的所有点的根节点是同一个。将每个点映射成一个数字。先假设每个点的根节点就是他们自己，然后我们以此输入连通的点对，然后将其中一个点的根节点赋成另一个节点的根节点，这样这两个点所在连通区域又相互连通了。 
并查集的主要操作有：

• find(int m)：这是并查集的基本操作，查找m的根节点。
• checkConnection(int m,int n)：判断m，n两个点是否在一个连通区域。
• Union(int m,int n):合并m，n两个点所在的连通区域。

当然，并查集还有较多优化，比如怎么避免并查集的树退化成一条链。合并的时候，应该尽量让小树根节点连在大树根节点上，尽量使得整个树扁平。

代码：

class Solution {    //union find主要找根节点，union 主要对两两（右下）进行合并，同时若一个是边缘的，则使另一个也为边缘private:    vector<int>root;    vector<bool>isedge;public:    void unions(int x, int y){        int root_x = find(x);        int root_y = find(y);        if(root_x == root_y) return;        root[root_x] = root_y; //将root_x 作为root_y的子节点        if(isedge[root_x]) isedge[root_y] = true;    }    int find(int x){        while(root[x] != x){            root[x] = root[root[x]];            x = root[x];        }        return x;    }    void solve(vector<vector<char>>& board) {        if(board.size() == 0 || board[0].size() == 0) return;        int height = board.size(), width = board[0].size();        int n = height* width;        root = vector<int>(n, 0);        isedge = vector<bool>(n, false);        //1. 初始化root和isedge数组        for(int i = 0; i < n; i++){            int x = i / width, y = i % width;            root[i] = i;            if((board[x][y] == 'O') && ((x == 0) || (x == height-1) || (y == 0) || (y == width-1))) isedge[i] = true;        }        //2.union         for(int i = 0; i < n; i++){            int x = i/width, y = i % width;            if((x < height -1) && (board[x][y] == board[x+1][y])) unions(i, i + width); //跟下面的进行合并            if((y < width -1) && (board[x][y] == board[x][y+1])) unions(i, i+1);         }        //3.find        for(int i = 0; i < n; i++){            int x = i/width, y = i%width;            if(board[x][y] == 'O' && !isedge[find(i)]) board[x][y] = 'X';        }    }};