LeetCode130：Surrounded Regions

题目：

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

解题思路：

首先在遍历最外面的四条边，如果遇到O，则表示有出路，这时，以找到的O点为起点，采用BFS或DFS进行遍历，找到与其他与该O点相邻的O点，然后将其置为*，第一步处理完后，再重新遍历整个矩阵，将为*的点置为O，为O的点置为X即可。

实现代码：

#include <iostream>

#include <vector>

#include <queue>

using namespace std;



/*

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

*/



class Solution {

public:

    void solve(vector<vector<char>> &board) {

        if(board.empty() || board[0].empty())

            return ;

        int rows = board.size();

        int cols = board[0].size();

        for(int i = 0; i < rows; i++)

        {

            if(board[i][0] == 'O')

                bfs(i, 0, board, rows, cols);

            if(board[i][cols-1] == 'O')

                bfs(i, cols-1, board, rows, cols);    

        }

        for(int j = 0; j < cols; j++)

        {

            if(board[0][j] == 'O')

                bfs(0, j, board, rows, cols);

            if(board[rows-1][j] == 'O')

                bfs(rows-1, j, board, rows, cols);

        }

        

        for(int i = 0; i < rows; i++)

            for(int j = 0; j < cols; j++)

                if(board[i][j] == 'O')

                    board[i][j] = 'X';

                else if(board[i][j] == '*')

                    board[i][j] = 'O';

  

    }

    

    void bfs(int i, int j, vector<vector<char>> &board, int rows, int cols)

    {

        queue<pair<int, int>> qu;

        qu.push(make_pair(i, j));

        while(!qu.empty())

        {

            pair<int, int> p = qu.front();

            qu.pop();

            int ii = p.first;

            int jj = p.second;

            if(ii < 0 || ii >= rows || jj < 0 || jj >= cols || board[ii][jj] != 'O')

                continue;

            board[ii][jj] = '*';

            qu.push(make_pair(ii, jj-1));

            qu.push(make_pair(ii, jj+1));

            qu.push(make_pair(ii-1, jj));

            qu.push(make_pair(ii+1, jj));

        

        }

    }

};



int main(void)

{

    return 0;

}