【leetcode】Surrounded Regions（middle）☆

Given a 2D board containing 'X' and '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

思路：

反向思考，先找到没有被包围的区域，标记为‘+’,再把标为‘+’的区域标为‘O',标为’O'的区域改为‘X'。没有被包围的区域一定是与最外圈的’O'相连的区域，所以要先遍历区域的上下左右边界，找到‘O'的地方。接下来的问题是如何把所有与最外层‘O'相连的区域标记上。有两种思路：BFS和DFS。

所谓BFS是指把当前标记位置的上下左右都标记一遍，然后再标记相邻点的上下左右位置。不需递归，用队列。

DFS是指把当前标记位置的向一个方向标记，比如一直向左，直到没有可标记的，再换一个方向。需要递归。

代码里面BFS可以通过，DFS栈溢出了。

void solve(vector<vector<char>> &board) {
        if(board.size() == ) return;
        int rowNum = board.size();
        int colNum = board[].size();
        //遍历最外面一圈，找‘O'
        //最上
        for(int j = ; j < colNum; j++)
        {
            if(board[][j] == 'O')
                BFS(board, , j);
        }
        //最下
        for(int j = ; j < colNum; j++)
        {
            if(board[rowNum - ][j] == 'O')
                BFS(board, rowNum - , j);
        }
        //最左
        for(int i = ; i < rowNum; i++)
        {
            if(board[i][] == 'O')
                BFS(board, i, );
        }
        //最右
        for(int i = ; i < rowNum; i++)
        {
            if(board[i][colNum - ] == 'O')
                BFS(board, i, colNum - );
        }

        for(int i = ; i < rowNum; i++)
        {
            for(int j = ; j < colNum; j++)
            {
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                if(board[i][j] == '+')
                    board[i][j] = 'O';
            }
        }

    }
    void DFS(vector<vector<char>> &board, int r, int c)
    {
        if(r >=  && c >=  && r < board.size() && c < board[].size() && board[r][c] == 'O')
        {
            board[r][c] = '+';
            DFS(board, r - , c);
            DFS(board, r + , c);
            DFS(board, r, c - );
            DFS(board, r, c + );
        }
    }
    void BFS(vector<vector<char>> &board, int r, int c)
    {
        queue<pair<int, int>> q;
        q.push(make_pair(r, c));
        while(!q.empty())
        {
            int i = q.front().first;
            int j = q.front().second;
            q.pop();
            if(i >=  && j >=  && i < board.size() && j < board[].size() && board[i][j] == 'O')
            {
                board[i][j] = '+';
                q.push(make_pair(i - , j));
                q.push(make_pair(i + , j));
                q.push(make_pair(i, j - ));
                q.push(make_pair(i, j + ));
            }

        }

    }

上面的代码已经是优化过的了，我自己写的时候只写出了DFS的，而且自己也没有意识到是DFS。代码也很繁琐。注意通过把判断条件放在一起来简化代码。

我原本很挫的代码：栈溢出。

class Solution {
public:
    void solve(vector<vector<char>> &board) {
        if(board.size() == ) return;
        int rowNum = board.size();
        int colNum = board[].size();
        //遍历最外面一圈，找‘O'
        //最上
        for(int j = ; j < colNum; j++)
        {
            if(board[][j] == 'O')
            {
                board[][j] = '+';
                mySolve(board, , j);
            }
        }
        //最下
        for(int j = ; j < colNum; j++)
        {
            if(board[rowNum - ][j] == 'O')
            {
                board[rowNum - ][j] = '+';
                mySolve(board, rowNum - , j);
            }
        }
        //最左
        for(int i = ; i < rowNum; i++)
        {
            if(board[i][] == 'O')
            {
                board[i][] = '+';
                mySolve(board, i, );
            }
        }
        //最右
        for(int i = ; i < rowNum; i++)
        {
            if(board[i][colNum - ] == 'O')
            {
                board[i][colNum - ] = '+';
                mySolve(board, i, colNum - );
            }
        }

        for(int i = ; i < rowNum; i++)
        {
            for(int j = ; j < colNum; j++)
            {
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                if(board[i][j] == '+')
                    board[i][j] = 'O';
            }
        }

    }
    void mySolve(vector<vector<char>> &board, int r, int c)
    {
        if(r -  >=  && board[r - ][c] == 'O') //上
        {
            board[r - ][c] == '+';
            mySolve(board, r - , c);
        }
        if(r +  < board.size() && board[r + ][c] == 'O') //下
        {
            board[r + ][c] == '+';
            mySolve(board, r + , c);
        }
        if(c -  >=  && board[r][c - ] == 'O') //左
        {
            board[r][c - ] == '+';
            mySolve(board, r, c - );
        }
        if(c +  < board[].size() && board[r][c + ] == 'O') //下
        {
            board[r][c + ] == '+';
            mySolve(board, r, c + );
        }
    }
};