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;
}