【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() == 0) return;
        int rowNum = board.size();
        int colNum = board[0].size();
        //遍历最外面一圈，找‘O'
        //最上
        for(int j = 0; j < colNum; j++)
        {
            if(board[0][j] == 'O')
                BFS(board, 0, j);
        }
        //最下
        for(int j = 0; j < colNum; j++)
        {
            if(board[rowNum - 1][j] == 'O')
                BFS(board, rowNum - 1, j);
        }
        //最左
        for(int i = 0; i < rowNum; i++)
        {
            if(board[i][0] == 'O')
                BFS(board, i, 0);
        }
        //最右
        for(int i = 0; i < rowNum; i++)
        {
            if(board[i][colNum - 1] == 'O')
                BFS(board, i, colNum - 1);
        }

        for(int i = 0; i < rowNum; i++)
        {
            for(int j = 0; 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 >= 0 && c >= 0 && r < board.size() && c < board[0].size() && board[r][c] == 'O')
        {
            board[r][c] = '+';
            DFS(board, r - 1, c);
            DFS(board, r + 1, c);
            DFS(board, r, c - 1);
            DFS(board, r, c + 1);
        }
    }
    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 >= 0 && j >= 0 && i < board.size() && j < board[0].size() && board[i][j] == 'O')
            {
                board[i][j] = '+';
                q.push(make_pair(i - 1, j));
                q.push(make_pair(i + 1, j));
                q.push(make_pair(i, j - 1));
                q.push(make_pair(i, j + 1));
            }
            
        }

    }

 

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() == 0) return;
        int rowNum = board.size();
        int colNum = board[0].size();
        //遍历最外面一圈，找‘O'
        //最上
        for(int j = 0; j < colNum; j++)
        {
            if(board[0][j] == 'O')
            {
                board[0][j] = '+';
                mySolve(board, 0, j);
            }
        }
        //最下
        for(int j = 0; j < colNum; j++)
        {
            if(board[rowNum - 1][j] == 'O')
            {
                board[rowNum - 1][j] = '+';
                mySolve(board, rowNum - 1, j);
            }
        }
        //最左
        for(int i = 0; i < rowNum; i++)
        {
            if(board[i][0] == 'O')
            {
                board[i][0] = '+';
                mySolve(board, i, 0);
            }
        }
        //最右
        for(int i = 0; i < rowNum; i++)
        {
            if(board[i][colNum - 1] == 'O')
            {
                board[i][colNum - 1] = '+';
                mySolve(board, i, colNum - 1);
            }
        }

        for(int i = 0; i < rowNum; i++)
        {
            for(int j = 0; 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 - 1 >= 0 && board[r - 1][c] == 'O') //上
        {
            board[r - 1][c] == '+';
            mySolve(board, r - 1, c);
        }
        if(r + 1 < board.size() && board[r + 1][c] == 'O') //下
        {
            board[r + 1][c] == '+';
            mySolve(board, r + 1, c);
        }
        if(c - 1 >= 0 && board[r][c - 1] == 'O') //左
        {
            board[r][c - 1] == '+';
            mySolve(board, r, c - 1);
        }
        if(c + 1 < board[0].size() && board[r][c + 1] == 'O') //下
        {
            board[r][c + 1] == '+';
            mySolve(board, r, c + 1);
        }
    }
};