130. Surrounded Regions

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本文介绍一种算法，用于在包含‘X’和‘O’字符的二维棋盘上捕获被‘X’包围的所有‘O’区域，并通过翻转这些区域内的字符实现捕获。该算法首先搜索棋盘边缘的‘O’字符，并将其以及相邻的‘O’字符标记为‘W’，最后将未被标记的‘O’字符变为‘X’。

问题描述
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 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
```
解决思路
对边缘的元素进行搜索，将边缘的o元素，以及与边缘o元素相邻的o元素变成‘w‘存储。然后将剩下的o元素变成x，将w重新变回o即可。值得注意的是，这里需要考虑空间问题，要进行枝，搜索过的就不要再搜索了。。。
代码

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int m = board.size();
        if (m == 0)
            return;
        int n = board[0].size();
        for (int i = 0; i < n; ++i) {
            helper(board,0,i,m,n);
            helper(board,m-1,i,m,n);
        }
        for (int i = 0; i < m; ++i) {
            helper(board,i,0,m,n);
            helper(board,i,n-1,m,n);
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'W')
                    board[i][j] = 'O';
                else if (board[i][j] == 'O')
                    board[i][j] = 'X';
            }
        }
    }
    void helper(vector<vector<char>>& board, int i, int j, int row, int col) {
        if (board[i][j] == 'O') {
            board[i][j] = 'W';
            if (i > 1)
                helper(board,i-1,j,row,col);
            if (j > 1)
                helper(board,i,j-1,row,col);
            if (i + 1 < row)
                helper(board,i+1,j,row,col);
            if (j + 1 < col)
                helper(board,i,j+1,row,col);
        }
    }
};