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.

    Example:

    1. X X X X
    2. X O O X
    3. X X O X
    4. X O X X

    After running your function, the board should be:

    1. X X X X
    2. X X X X
    3. X X X X
    4. X O X X

    Explanation:

    Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.

    题意

    在棋盘上找到所有被’X’完全包围的’O’，将其翻转为’X’(类似围棋)。

    思路

    如果’O’在四条边上出现，那么与这个’O’相连的所有’O’都不可能被’X’包围。从这一点入手，我们可以先找到所有在边上出现的’O’，dfs处理将所有相连的’O’都先翻转为’Q’；重新遍历二维数组，剩余的’O’都需要被翻转为’X’，而遇到’Q’则将其重新翻转为’O’。

    代码实现

    1. class Solution {
    2.     private int m, n;
    4.     public void solve(char[][] board) {
    5.         if (board.length == 0 || board[0].length == 0) {
    6.             return;
    7.         }
    9.         m = board.length;
    10.         n = board[0].length;
    12.         // 从四边入手dfs处理
    13.         for (int i = 0; i < n; i++) {
    14.             if (board[0][i] == 'O') {
    15.                 dfs(board, 0, i);
    16.             }
    17.             if (board[m - 1][i] == 'O') {
    18.                 dfs(board, m - 1, i);
    19.             }
    20.         }
    21.         for (int i = 0; i < m; i++) {
    22.             if (board[i][0] == 'O') {
    23.                 dfs(board, i, 0);
    24.             }
    25.             if (board[i][n - 1] == 'O') {
    26.                 dfs(board, i, n - 1);
    27.             }
    28.         }
    30.         // 处理剩余被包围'O'以及还原'Q'
    31.         for (int i = 0; i < m; i++) {
    32.             for (int j = 0; j < n; j++) {
    33.                 if (board[i][j] == 'O') {
    34.                     board[i][j] = 'X';
    35.                 } else if (board[i][j] == 'Q') {
    36.                     board[i][j] = 'O';
    37.                 }
    38.             }
    39.         }
    40.     }
    42.     private void dfs(char[][] board, int i, int j) {
    43.         if (i < 0 || j < 0 || i >= m || j >= n || board[i][j] != 'O') {
    44.             return;
    45.         }
    47.         board[i][j] = 'Q';
    48.         dfs(board, i - 1, j);
    49.         dfs(board, i + 1, j);
    50.         dfs(board, i, j - 1);
    51.         dfs(board, i, j + 1);
    52.     }
    53. }